If $X$ is first countable, is $f(X)$ also first countable? I'm trying to understand the proof of this theorem: $X$ is first countable and $f:X\rightarrow Y$ is a function. Then $f$ is continuous at $x\in X$ iff for every sequence $(x_n)$ which converges to $x$, $f(x_n)$ converges to $f(x)$.
Fix $x\in X$. To prove the "only if" statement, it suffices to show that $f(\overline{A})\subseteq\overline{f(A)}$ for each $A\subset X$ such that $x\in\overline{A}$.
Since $x\in\overline{A}$, by a previously proven lemma, there is a sequence $(x_n)$ in $A$ which converges to $x$. By assumption, the sequence $f(x_n)$ in $Y$ converges to $f(x)$. Since $f(x_n)\in f(A)$ for each $n\in\mathbb{N}$, we have $f(x)\in\overline{f(A)}$.
The last line of the proof seems to imply that if $X$ is first countable, then so is $f(X)$. But why is this true? It think it is only true if $f$ is continuous at $x$, but obviously we can't assume that.

Here is a continuous counter example to the question of the title.
Let Y be any space that is not first countable.
Let X be the points of Y with the discrete topology.
Let f be the the identity map from X to Y.  
To show f is continuous iff f preserves sequence limits use
the fact that f is continuous iff for all A,
$f(cl A) \subseteq cl f(A)$ and the theorem  K is closed
iff for all x, (x in K iff some sequence within K converges to x). 
cl = closure.
