Is this alternative notion of continuity in metric spaces weaker than, or equivalent to the usual one? I will try to be as clear as possible. 
For simplicity I will assume that the function $f$ for which we define continuity at some point is real function of a real variable $f: \mathbb R \to \mathbb R$, although the same line of reasoning should be the same even if we talk about continuity of a function at a point that is defined at some metric space and that has values in some metric space. To define continuity of a function $f$ at some point $x_0$ from its domain we have the following standard and famous definition, the $\varepsilon$-$\delta$ definition of continuity, which goes like this:

Definition 1: $f$ is continuous at the point $x_0$ from its domain if for every $\varepsilon>0$ there exists $\delta>0$ such that when $|x-x_0|<\delta$ then $|f(x)-f(x_0)|<\varepsilon$. 

It is clear that we could write $\delta (\varepsilon)$ instead of $\delta$ because there really is dependence of $\delta$ on $\varepsilon$.
Now, I was thinking about alternative definition of continuity that would go like this:

Definition 2: $f$ is continuous at the point $x_0$ from its domain if there exist two sequences $\varepsilon_n$ and $\delta_n$ such that for every $n \in \mathbb N$ we have $\varepsilon_n>0$ and $\delta_n>0$ and $\lim_{n\to\infty}\varepsilon_n=\lim_{n\to\infty}\delta_n=0$ and when $|x-x_0|<\delta_n$ then $|f(x)-f(x_0)|<\varepsilon_n$.

We could also write here $\delta_n(\varepsilon_n)$ instead of $\delta_n$ because obviously there is dependence of $\delta_n$ on $\varepsilon_n$. 
It is clear that second definition does not require that for every $\varepsilon$ there exist $\delta(\varepsilon)$ (which includes in itself an uncountable number of choices for $\varepsilon$) but instead requires that for every member of the sequence $\varepsilon_n$ there is some $\delta_n$ (and this includes in itself only countable number of choices because the set $\{\varepsilon_n : n \in \mathbb N\})$ is countable).
It is clearly obvious that definition 1 implies definition 2, and the real question is is the converse true, in other words:

If the function is continuous at some point according to definition 2, is it also continuous at the same point according to definition 1?

 A: The two definitions are equivalent in metric spaces because metric spaces are first countable: every point in a metric space has a countable neighborhood basis. (We'll explain that further below, when discussing Definition 2.)
The standard Definition 1. rephrases the definition from general topology of continuity at a point into a more concrete form that takes advantage of what $\Bbb R$ has to offer. The general definition, shown in three equivalent forms, is as follows:
$$\begin{align}
&\it\text{($\forall$ neighborhoods $V$ of $\,f(x)$)$\,$ $f^{-1}(V)$ is a neighborhood of $x$}\\
&\it\text{($\forall$ neighborhoods $V$ of $\,f(x)$)$\,$ ($\exists$ a neighborhood $U$ of $x$) $\:U \subseteq f^{-1}(V)$}\\
&\it\text{($\forall$ neighborhoods $V$ of $\,f(x)$)$\,$ ($\exists$ a neighborhood $U$ of $x$) $\:f(U) \subseteq V$}\\
\end{align}$$
Without loss of generality, Definition 1. restricts attention not just to open neighborhoods but to the basic open neighborhoods, the open balls centered at $x$ and $f(x)$. In $\Bbb R$, the open balls around $u$ are all the intervals $I_{u,r} = (u-r, u+r), r\in\Bbb R_+$. Given $x$ and $f(x)$, these neighborhoods are specified by their radii, so the definition becomes:
$$
\forall \varepsilon\,\exists\delta\,f(I_{x,\delta}) \subseteq  I_{f(x),\varepsilon},$$
which is to say,
$$
\forall \varepsilon\,\exists\delta\,\forall y\,(y\in I_{x,\delta}\to f(y)\in I_{f(x),\varepsilon}).
$$
But note that $v\in I_{u,r}$ exactly when $\lvert v-u\rvert < r$, so we can recast the previous form into the standard definition:
$$
\forall \varepsilon\,\exists\delta\,\forall y\,(\lvert y-x\rvert < \delta \to \lvert f(y)-f(x)\rvert < \varepsilon).
$$
This form is much friendlier to computation and approximation.
Definition 2. takes advantage of the fact that metric spaces are first countable: for every $x$ in a metric space $X$, there is a countable collection $(U_{x,n})_{n\in\Bbb N}$ of neighborhoods of $x$ such that any neighborhood $V$ of $x$ contains some $U_{x,n}$. Such a collection is called a (countable) neighborhood basis at $x$. The sequences $\varepsilon_n$ and $\delta_n$ give countable neighborhood bases for $f(x)$ and $x$ respectively:

*

*$U_{x,n} = I_{x,\delta_n}$,

*$U_{f(x),n} = I_{f(x),\varepsilon_n}$.

This form is even more amenable to computation. Suppose the  sequence $(\varepsilon_n)$ is given effectively: there is a procedure which on input $n$ produces a rational $\varepsilon_n$. For many $f$ we can devise another procedure which on inputs $n$ and $(\varepsilon_i)_{i\le n}$ produces $\delta_n$. (Think of the expressions which arise in analysis proofs that establish continuity of specific functions, e.g. "take $\delta = \sqrt 2 \varepsilon/K$".) In general, computable $f$ aren't just continuous at a point, and we may even be able to give a procedure that computes a modulus of continuity for $f$.
Finally, there's a third possible definition, in a sense intermediate between Definitions 1. and 2. and also equivalent to them:

$f$ is continuous at $x$ iff for every sequence of positive reals $(\varepsilon_n)_{n\in \Bbb N} \to 0$, there is a sequence of positive reals $(\delta_n)_{n\in \Bbb N}$ such that for all $n$ and for all $y$, if $\lvert y-x\rvert < \delta_n$ then $\lvert f(y)-f(x)\rvert < \varepsilon_n$.

In more general topological spaces, which might not be metric spaces or even first countable, continuity can't be completely characterized by countable sequences. However, it can be characterized with the more general notion of a net. For a great self-contained examination of the limits (so to speak) of sequential convergence and the versatility of nets, see Pete Clark's notes on Convergence.
A: Indeeed definition $2$ implies definition $1$:
Let us assume that $f$ satisfies definition $2$, and try to prove it is continuous by definition $1$.
Let $\epsilon > 0$. Then, because $\lim_{n\to\infty} \epsilon_n = 0$, there exists such a $N$ that $\epsilon_N < \epsilon$.
Now, let us set $\delta=\delta_N$, and let $|x-x_0|<\delta$. Because $\delta=\delta_N$, this means that $|x-x_0|<\delta_N$, and by definition $2$, this means that $|f(x)-f(x_0)|<\epsilon_N$.
This further means that $|f(x)-f(x_0)|<\epsilon_N<\epsilon$, meaning that $f$ is continuous at $x_0$.
A: The second definition may not require that a $\delta$ exists for every $\epsilon > 0$, but we still get this as a consequence.  This is because if $\delta$ works for a given $\epsilon$, then $\delta$ will work for any larger $\epsilon$, too.  So, $\delta_{n}$ will work for every $\epsilon \geq \epsilon_{n}$.  Since $\epsilon_{n} \to 0$, then your second "definition of continuity" implies the original.
A: Definition 2 indeed implies Definition 1,
However:
There is a superfluous requirement in Definition 2. 
Namely, the requirement the $\delta_n\to 0$ is not needed. 
The most important part in the Definition of continuity is the fact that $\varepsilon>0$ can be as small as possible. On the other hand, apart from the fact that $\delta$ depends on $\varepsilon$, the size of $\delta$ is irrelevant.
So, it could be reworded as follows:

Definition 2. The function $f:A\to\mathbb R$ is continuous at $x_0\in A$ if there exist two sequences $\{\varepsilon_n\}_{n\in\mathbb N}$ and $\{\delta_n\}_{n\in\mathbb N}$ of positive numbers, with  $\lim_{n\to\infty}\varepsilon_n=
0$, such that, whenever $\,\lvert x-x_0\rvert<\delta_n$, then $\,\lvert\,f(x)-f(x_0)\rvert<\varepsilon_n$.

