An unorthodox way to find cardinalities? Let $\mathfrak{c}$ denote the cardinality of the continuum. I sketch an intuitive but non-rigorous argument that $|\mathbb{R}^\mathbb{N}| = \mathfrak{c}$, with the question:
Question: can this argument be made rigorous?

Sketch "proof": Let $f : \mathbb{R} \to \mathbb{R}, a \in \mathbb{R}$. It is a standard result that $f$ is continuous at $a$ iff:

*

*$\forall\varepsilon>0 \; \exists \delta>0$ s.t. $|x-a| < \delta \implies |f(x)-f(a)|  < \varepsilon$

*$\forall (x_n)$ s.t. $x_n \to a$, $f(x_n) \to f(a)$
(1) requires something to be true for every element of a set of cardinality $\mathfrak{c}$, while (2) requires something to be true for every element of:
$$S := \{(x_n) | x_n \to a\}$$
But since (1) and (2) are equivalent, we may deduce that $|S| = \mathfrak{c}$. It follows that $|\mathbb{R}^\mathbb{N}| = \mathfrak{c}$.
(This argument has inaccuracies, at least some of which can be fixed with observations made below.)

General idea: If two logical statements are equivalent and each "depends"(?) on sets of cardinality $C$ and $C'$ respectively, then we would expect that $C = C'$.
I outline two potential issues that I spotted with this type of reasoning below.

Objection A: Suppose we let $T$ = $\{S\}$ i.e a set with $S$ as an element. Then, we can rephrase (2) as:


*For every $s \in t$ of every $t \in T$, $f(s) \to f(a)$ (where $f(s)$ is interpreted in the obvious way).

But then $T$ is a set of cardinality $1$, and there's an issue because  $\mathfrak{c} \neq 1$. I think this issue could be remedied with a more rigorous approach, but I don't know any of the set theory which I expect is needed to do so.

Objection B: Consider the following two statements. We have that $x \le 0$ iff:


*$x \le a$, $\; \forall a \in A_1$ where $A_1 = \{0\}$

*$x \le a$, $\; \forall a \in A_2$ where $A_2 = \mathbb{R}^+ \cup \{0\}$
Then, (4),(5) are equivalent, but depend on sets of wildly different cardinalities $1 \neq \mathfrak{c}$ again.
The solution to this is a bit more obvious. It is clear that most of $A_2$ is redundant, so we could argue that it has "effective cardinality" $1$, since it suffices to know simply whether or not $x \le a$ for $a=0 \in A_2$. But that doesn't solve the issue of (6):


*$x < a$, $\; \forall a \in A_3$ where $A_3 = \mathbb{R}^+$
which is also equivalent to (4),(5) but appears to have countably infinite "effective cardinality". (Notably, $\mathbb{R}^+$ is not closed; I think that is key here.)

I expect that for both (A) and (B), there would need to be some kind of condition on what "types" of elements are allowed for the sets in question, and also the types of sets allowed.
Is there a way to reconcile all of these issues and make this a valid direction of argument? Is there any truth to the general idea described above, and if not, is there a clear error that can be pinpointed?
 A: The proof is completely invalid for the reason as Arturo Magidin has noted, that it would require that the effective cardinality and cardinality of a set is always equal, for which I even gave counterexamples in the original post.
Though unlikely, however, it may be possible to say more about the effective cardinalities described in the question -- it may be possible to equate them sometimes. Since this was not treated rigorously in the body of the question, I attempt a definition below:
Let $S$ be a set. Let $\rho : S \to \{0,1\}$ be a property. Then, let $T \subseteq S$. Define the deductive closure of $T$ (with respect to $\rho$) to be:
$$T' = \{t' \in S: \rho (t') \text{ can be inferred from knowing } \rho(t) \; \forall t \in T\}$$
So certainly $T \subseteq T'$. Now define the effective cardinality of $S$ to be the minimal cardinality of a set $T \subseteq S$ such that $T' = S$ (again, this depends on $\rho$, of course, which may itself depend on other parameters).
Example 1
Let $S = \mathbb{N}$ and $\rho : S \to \{0,1\}$, $p \in \mathbb{N}$ be fixed, but unknown.
$$\rho(x) = \mathbb{I}(p | x)$$
Then, the deductive closure of $\{x\}$ contains $\{nx : n \in \mathbb{N}\}$. However, no finite set has deductive closure equal to $S$, since for any finite set, there will exist two different values of $p$ which have the same values on this set. Thus, the effective cardinality of $S$ is $\aleph_0$. (Note that the effective cardinality of a set is bounded above by its cardinality.)
Example 2
Let $S = \mathbb{N}$ and $\rho : S \to \{0,1\}$, $k \in \mathbb{N}$ be fixed (but again this does not mean known), $n \in \mathbb{N}$ known.
$$\rho(x) = \mathbb{I}(x \equiv k \text{ mod } n)$$
Then, the effective cardinality of $S$ is $n$.
Example 3
Let $S = \mathbb{N}$ and $\rho : S \to \{0,1\}$,
$$\rho(x) = \mathbb{I}(x \text{ is composite})$$
Then, the deductive closure of any subset of $S$, even the empty set, is $S$ because we do not need to know ex-ante that a particular integer is prime in order to show that the composite naturals are exactly $\mathbb{N} \setminus \{2,3,...\}$ -- this can be done "from the ground up"; from the standard axioms. So $S$ has effective cardinality $0$.
Example 4
Let $S = \{(y_n) : y_n \to a \}$ and $\rho : S \to \{0,1\}$, $f:\mathbb{R} \to \mathbb{R}$ be fixed.
$$\rho(x) = \mathbb{I}(f(x) \to f(a)) $$
where $f(s)$ is the sequence obtained by applying $f$ to $x \in S$ term-wise.
Let $T \subseteq S$. Then, the deductive closure of $T$ is every sequence tending to $a$ (i.e. sequences in $S$) that consists of terms which are elements of sequences in $T$. Suppose $T$ is countable. Then, we do have that $T'$ is uncountable, but it is never the case that $T' = S$. So the effective cardinality of $S$ is $\mathfrak{c}$.
This is not necessarily true if $S = \{(y_n) : y_n \to a, y_n \in \mathbb{Q} \}$. As noted, it can be the case that a countable set has uncountable deductive closure, so the fact that $|\mathbb{Q}^{\mathbb{N}}| = \mathfrak{c}$ is not necessarily an issue. It is also necessary to consider Objection A, which has not been addressed by the other answer.

Following from Objection A, it is clear that certain set-theoretic conditions may be required; otherwise, assuming the result, we easily obtain falsehoods such as $\mathfrak{c}= \aleph_0$ or $\mathfrak{c}=1$, which I cheerfully demonstrated myself. There must be a distinction in some way between sets, and sets which contain sets. Admittedly, this is vague, but again, I do not understand set theory.
This consideration is crucial if attempting to accurately determine the effective cardinality of a set. Without such a measure, it would be futile to carry out this argument. I have bothered to compute effective cardinalities above because the reader will retrospectively notice that in Examples 1-3, all sets $S$ had elements which do not themselves contain elements (again, forgive any set-theoretic naivety here), unlike sequences or intervals in $\mathbb{R}$, the latter of which have uncountably many elements themselves.
(And in the first part of Example 4, this only tells us further that the effective cardinality is at  least $\mathfrak{c}$, which we already know to be an upper bound. So this does not affect the conclusion that the effective cardinality is $\mathfrak{c}$.)
Thus, any true answer should first address Objection A, bearing in mind this was always the crux of the question ("can the reasoning be fixed?"). Otherwise, it will only re-tread territory already covered multiple times in the question and comments.
A: As I understand the general shape of the argument, it runs as follows:

We have two sets, $R$ and $S$. We wish to show that $R$ and $S$ have the same cardinality.
We find a set $A$ and a nonempty subset $B$, and properties $P$ and $Q$ such that:


*

*For all $x\in A$, $\Bigl(x\in B\iff \bigl(\forall r\in R (P(x,r))\bigr)\Bigr)$;

*For all $x\in A$, $\Bigl(x\in B\iff \bigl(\forall s\in S (Q(x,s))\bigr)\Bigr)$;


both hold.
We conclude that $R$ and $S$ have the same cardinality.

You seem to sketch such an argument with $A$ the set of all function $\mathbb{R}\to \mathbb{R}$, $B$ the set of functions in $A$ that are continuous at $a$, $R=\mathbb{R}_{\gt 0}$ (or some other subset of the positive reals of cardinality $\mathfrak{c}$), $S$ the set of all sequences of real numbers; for $f\in A$ and $\epsilon\in R$, $P(x,\epsilon)$ is "$\exists \delta\gt 0$ such that if $|x-a|\lt \delta$ then $ |f(x)-f(a)|\lt \epsilon$"; and for $f\in A$ and $(x_n)\in S$, $Q(f,(x_n))$ is "if $x_n\to a$, then $f(x_n)\to f(a)$".
Now, as written, you acknowledge that the argument is not valid, as you can produce specific instances of $A$, $B$, $R$, $S$, $P$, and $Q$ in which the cardinalities of $R$ and $S$ are distinct. For example:


*

*$A=\mathbb{R}$, $B=\{x\in\mathbb{R}\mid x\lt 0\}$, $R=\{0\}$, $S=[0,\infty)$, $P(x,r)="x\lt r"$, and $Q(x,s) = "x\lt s$.

*$A=\mathbb{R}$, $B=\{x\in\mathbb{R}\mid x\lt 0\}$, $R=\{0\}$, $S=(0,\infty)$, $P(x,r)="x\lt r"$, and $Q(x,s)="x\lt s$".


You then suggest that perhaps the problem in those specific examples is that the sets are somehow redundant. From comments, it would seem that you would then want to define something that I interpreted as going along the following lines (I will use a different term, since I am told what I interpreted is not what was intended):

For sets $A$, $B$, $R$ and a property $P$ such that
$$x\in B\iff \forall r\in R (P(x,r))$$
we say a subset $R'$ of $R$ "determines for $B$", or is an "STD" (subset that determines) if $B$ is understood from context, iff
$$x\in B\iff \forall r\in R' (P(x,r))$$
also holds. Define the "STD-cardinality" of $R$ (the "subset that determines") to be $\min\{ |R'|\mid R'\text{ is a subset of }R\text{ that determines for }B\}.$

(Note that we cannot talk about a "smallest subset of $R$ that determines for $B$ in general, since there may not be a least such subset; however, any nonempty set of cardinals has a least element, so we can talk about the minimum of the cardinals of all STDs).
Then the original argument would be modified as follows:

We have two sets, $R$ and $S$. We wish to show that $R$ and $S$ have the same cardinality.
We find a set $A$ and a nonempty subset $B$, and properties $P$ and $Q$ such that:


*

*For all $x\in A$, $\Bigl(x\in B\iff \bigl(\forall r\in R (P(x,r))\bigr)\Bigr)$;

*For all $x\in A$, $\Bigl(x\in B\iff \bigl(\forall s\in S (Q(x,s))\bigr)\Bigr)$;


both hold.
We conclude that $R$ and $S$ have the same STD-cardinality.

Note that the roles of $R$ and $S$ are in symmetric and interchangeable, so if we are going to replace the cardinality of $S$ with its STD-cardinality, then we must also replace the cardinality of $R$ with its STD-cardinality in the conclusion; note also that even if this particular argument held, it would not work for the instance you are trying to establish, since you are trying to conclude that $|R|=|S|$, so you would need to show that both specific sets you have are equinumerous with their STD-cardinality, which as I show below is not the case for $R$.
This also does not hold, as you yourself note, since in the case of $A=\mathbb{R}$, $B=\{x\in\mathbb{R}\mid x\lt 0\}$, $R=\{0\}$, $S=(0,\infty)$, and $P(x,r)=Q(x,r) = "x\lt r"$, the STD-cardinality of $R$ is clearly $1$, while the STD-cardinality of $S$ must be strictly larger.
So the argument cannot hold in the abstract, despite the "general idea" described as "reasonable." It may appear reasonable, but it is fallacious and does not work.
So this general idea just doesn't work; not in the abstract.
So perhaps, it can still be made to work in the specific instance of $A=\{f\colon \mathbb{R}\to\mathbb{R}\}$, $B=\{f\in A\mid f\text{ is continuous at }a\}$, $R=(0,\infty)$, $S$ the set of all real sequences, $P(f,r)$ being "there exists $\delta\gt 0$ such that if $|x-a|\lt \delta$ then $|f(x)-f(a)|\lt r$", and $Q(f,(x_n))$ being "if $x_n\to a$ then $f(x_n)\to f(a)$". So for this and only this situation, dealing with functions in the reals.
But, no, it doesn't work in this individual instance either.
First:
Lemma. A subset $R'$ of $(0,\infty)$ has the property that for all $f\in A$,
$$f\in B\iff \forall r\in R'(P(f,r))$$
if and only if $\inf(R')=0$. In particular, $R'$ must be infinite, and there are countable subsets $R'$ that have the property.
Proof. Assume $\inf(R')=0$.
Let $f\in B$. I claim that $\forall r\in R'(P(f,r))$ if and only if $\forall r\in R(P(f,r))$. Indeed, clearly if $P(f,r)$ holds for all $r\in R$ then it holds for all $r\in R'$, since $R'\subseteq R$. Now suppose that for all $r\in R'$, $P(f,r)$ holds. Let $r\in R$. Since $\inf(R')=0\lt r$, there exists $r'\in R'$ such that $r'\lt r$. Since $P(f,r')$ holds, let $\delta\gt 0$ be such that if $|x-a|\lt\delta$ then $|f(x)-f(a)|\lt r'$. Then $\delta$ also suffices for $r$, since $|f(x)-f(a)|\lt r'\lt r$ holds. Thus, $P(x,r)$ holds as well. This establishes the claim.
Thus, if $\inf(R')=0$, then the following are equivalent:

*

*$f\in B$;

*$\forall r\in R (P(f,r))$

*$\forall r\in R' (P(f,r))$.

Conversely, suppose that $\inf(R')=\epsilon_0\gt 0$. Consider the function $f$ with defined as $f(x)=0$ for all $x\neq a$, and $f(a)=\frac{\epsilon_0}{2}$. Then $f$ is not continuous at $a$, so $f\notin B$. However, $P(f,r)$ holds for all $r\in R'$. Thus, $R'$ does not determine membership in $B$.
This establishes that a subset $R'$ of $R$ determines membership in $B$ via property $P$ if and only if $\inf(R')=0$. Since $R$ does not contain $0$, no finite subset can have infimum $0$; and the set $\{\frac{1}{n}\mid n\in\mathbb{N}\}$ is a countable subset of $R$ with infimum $0$. $\Box$
So the STD-cardinality of $R$ here is $\aleph_0$. In particular, it does not equal the cardinality of $R$. This already dooms this attempt at establishing $|R|=|S|$.
On the other hand,
Lemma. Let $S'$ be a subset of $S$ such that for all $f\in A$,
$$f\in B\iff \forall (x_n)\in S'(Q(f,(x_n))).$$
Then $S'$ is not countable.
Proof. We show that for any countable subset $S'$ of $S$, there is a function $f$ that is not continuous at $a$ but for which $Q(f,(x_n))$ holds for every $(x_n)\in S'$.
Let $S'$ be a countable subset of $S$, and let $X=\{r\in\mathbb{R}\mid \exists (x_n)\in S', n\in\mathbb{N}\text{ such that }r=x_n\}$. That is, all numbers that appear as a term in a sequence in $S'$. Now let $f$ to be the indicator function for $X\cup\{a\}$: $f(x)=1$ if $x\in X\cup\{a\}$, and $f(x)=0$ if $x\notin X\cup\{a\}$. Note that for all $(x_n)\in S'$, for all $n\in\mathbb{N}$, $f(x_n)=1$, and $f(a)=1$. Thus, for all $(x_n)\in S'$, we have $f(x_n)\to f(a)$, whether or not $x_n\to a$. That is, for all $(x_n)\in S'$ we have $Q(f,(x_n))$.
However, $Q(f,(x_n))$ does not hold for all $(x_n)\in S$. Indeed, define a sequence $(x_n)$ as follows: given $n\in\mathbb{N}$, pick $x_n\in (a-\frac{1}{n},a+\frac{1}{n})\setminus X$; that is, pick a real number which is not in $X$ and is within $\frac{1}{n}$ of $a$. Such a number must exist, since $(a-\frac{1}{n},a+\frac{1}{n})$ is uncountable, but $X$ is countable, so the difference is not empty. Thus, $(x_n)\in S$, $x_n\to a$, but $f(x_n)=0$ for all $n\in\mathbb{N}$; however, $f(a)=1$. Hence, $Q(f,(x_n))$ does not hold. $\Box$
This means that the STD-cardinality of $S$ is strictly larger than $\aleph_0$. Using a similar argument we can show that no subset of cardinality strictly smaller than $\mathfrak{c}$ can suffice. (Since we are doing thisin order to determine the cardinality of $S$ in the first place, this is as much as we can say here). Regardless, the STD-cardinality of $R$ is not equal to the STD-cardinality of S, even though both sets characterize continuity. So the "general ideal" does not hold in the particular circumstance either, even after attempting to remove redundancies from the sets in question.
