Cardinallity of the set of continuous functions $X \to Y$, when space $X$ is separable, $Y$ is Hausdorff and $|Y|\leq 2^{\aleph_0}$. Problem statement:

Show that if $X$ is a separable space and $Y$, and $|Y|\leq 2^{\aleph_0}$, then there are at most $2^{\aleph_0}$ continuous functions from $X$ to $Y$.

It's not mentioned, but it'll be safe to assume that $Y$ is intended to be $T_2$. Otherwise, $X=\mathbb{R}$ with the standard topology and $Y=\{0,1\}$ with the trivial topology, would be a counter example.
Now, it can be show that if $f,g:X\to Y$ are continuous and share the same values on a dense subset, then $f=g$.
Also, if $Y$ is separable, then we can take some dense countable subset $D_Y \subseteq Y$, so for any point $y \ \in Y$ we have: $$\{q_n\}_{n=1}^{\infty}, (q_n \in D_Y) \\ \lim_{n \to \infty}q_n =y $$ so any function $f: X \to Y$ can be show to be equal to the to the limit of some series $\{f_n\}_{n=1}^{\infty}$ with $\text{Im}f_n \subseteq D_Y $.
Thus we get, in the case $Y$ is separable, $D_x \subseteq X $ is dense and countable, and $A$ is the set of continuous functions $X \to Y$:
$$ |A| = |(D_Y^{D_X})^{\mathbb{N}}| \leq (\aleph_0^{\aleph_0})^{\aleph_0} = 2^{\aleph_0} $$
Problem / Question:

$Y$ is not known to be separable. It is true that any continous function will map $X$ to a separable subset of $Y$, but how can it be shown that there are  at most $2^{\aleph_0}$ such subpaces?

 A: Let $Z\subset X$ be countable and dense. Since both $X$ and $Y$ are $T_2$, a continuous function $f:X\to Y$ is completely determined by its values on $Z$ (limits of sequences commute with $f$ and a sequence can have at most one limit in $Y$). Therefore, there are at most
$$|Y|^{|Z|} \le (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0}$$
continuous functions from $X$ to $Y$, with no need of assuming separability of $Y$.

As a fun, but completely useless remark, notice that if $X$ is uncountable, $|Y|\ge2$ and if we assume the continuum hypothesis, then this proves non-constructively that there exist non-continuous functions from $X$ to $Y$.
A: $Y$ is indeed intended to be Hausdorff. In that case we can show (but not using sequences, if we are in general spaces, not first countable ones) that if $f = g$ on $D$, a (countable) dense subset of $X$, then $f=g$ on $X$. $(\Delta_Y \subseteq Y \times Y$ is closed, where $\Delta_Y = \{(y,y): y \in Y\}$, precisely when $Y$ is Hausdorff, and the map $(f,g): X \to Y \times Y, (f,g)(x) = (f(x), g(x))$ is continuous whenever $f$ and $g$ are. Then $D \subseteq (f,g)^{-1}[\Delta_Y]$ by assumption, and so $\overline{D} = X \subseteq (f,g)^{-1}[\Delta_Y]$ as well, so for all $x \in X$, $f(x) = g(x)$).
So the function $R$ of restriction from $C(X,Y)$ to $C(D,Y)$ is injective, so $$
|C(X,Y)| \le |C(D,Y)| \le |Y|^{|D|} \le (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0 \cdot \aleph_0} = 2^{\aleph_0}$$
as required. Separability of $Y$ is irrelevant. You are right that each individual $f : X \to Y$ that is continuous, $f[X]$ is separable , so that if $Y$ is not, we cannot have surjective continuous functions.
A: Theorem. Let $Y$ be Hausdorff and $f:X\to Y,\;g:X\to Y$ be continuous . Then $\{x\in X: f(x)\ne g(x)\}$ is open in $X.$ 
Corollary: If $f:X\to Y$ and $g:X\to Y$ agree on a dense subset of $X$, and $Y$ is Hausdorff, then $f=g.$
The theorem is easy: For $f(x)\ne g(x)$ let $f(x)\in  V_f$ and $g(x)\in V_g$ where $V_f,V_g$ are disjoint open subsets of $Y.$ Then $\forall x'\in (f^{-1}V_f)\cap (g^{-1}V_g)\;(f(x')\ne g(x')).$ 
Let $D$ be a dense subset of $X$ and let $Y$ be Hausdorff. Let $C(X,Y)$ be the set of continuous members of $Y^X.$ By the corollary above, the map $f\to f|_D$ is injective from $C(X,Y)$ to $Y^D,$ so $|C(X,Y)|\leq |Y^D|.$ 
If $|Y|\leq 2^{\omega}$ and $|D|\leq \omega$ then $|Y^D|=|Y|^{|D|}\leq (2^{\omega})^{\omega}=2^{(\omega\cdot\omega)}=2^{\omega}.$
Reference topic: The Jones Lemma.
