Separable topology and continuous functions 
Let $X$ be a separable topology and $Y$ a topology such that $|Y|\le 2^{\aleph_0}$.
Prove that there are at most $2^{\aleph_0}$ continuous functions from $X$ to $Y$.

I don't even know how to approach this kind of question...
 A: This is false if $Y$ is not assumed to be Hausdorff:
Denote as usual $2^{\aleph_0}$ by $\mathfrak{c}$ (continuum).
Let $X$ be the reals in the usual topology, and let $Y$ be the reals in the indiscrete topology. Then all functions into $Y$ are continuous and there are $\mathfrak{c}^\mathfrak{c} = 2^{\mathfrak{c}}> \mathfrak{c}$ such functions.
We can also use $Y$ as the cofinite topology on the reals and note that at least all bijections on the reals are continuous (and there are also $\mathfrak{c}^\mathfrak{c}$ many of those). This is a $T_1$ but not $T_2$ space, showing we need some assumption on $Y$.
If $Y$ is Hausdorff: pick $D$ dense and countable in $X$.
Then the map $f \to f|D$ from a continuous function from $X$ to $Y$ to all functions from $D$ to $Y$, is 1-1 (two functions that agree on a dense set, agree on the whole domain; this uses that $Y$ is Hausdorff). The image set of functions is $Y^D $ which has at most $$(2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0 \cdot \aleph_0} = 2^{\aleph_0}$$
many functions in it. As the restriction map is 1-1, the upperbound has been shown.
