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.

Firstly: $f[X]$ is not necessarily first countable, e.g. take $X = \mathbb{R}$ in the usual topology and let $\sim$ be the equivalence relation with classes $\{x\} , x \notin \mathbb{Z}$ and $\mathbb{Z}$ (We identify the integers to a point). Then the quotient space $Y = X/\sim$ in the quotient topology, induced by the standard map $q$ mapping $x$ to its class, is a continuous image (of $q$) which is not first countable at the class/point $[\mathbb{Z}]$.
Following your proof, which is quite correct: we assume $f$ preserves sequence limits. Then we want so show that for any $A \subseteq X$, $f[\overline{A}] \subseteq \overline{f[A]}$, which is indeed is one of the characterisations of continuity.
So let $f(x) \in f[\overline{A}]$, so $x \in \overline{A}$, which implies that there is a sequence $x_n$ from $A$ that converges to $x$ (this holds in first countable spaces! this is where we use it). Then $f(x_n) \to f(x)$ by the assumption on $f$, and as all $f(x_n) \in f[A]$, $f(x) \in \overline{f[A]}$ as required.
The last fact is true in all spaces: if $x_n \in B$ and $x_n \to x$ in the space, then $x \in \overline{B}$: let $O$ be any open neighbourhood of $x$. Then $O$ contains all $x_n$ for $n \ge N$ for some $N$. But in particular: $x_N \in O \cap B \neq \emptyset$, so $x \in \overline{B}$. But for the reverse (that being in the closure means that we have a sequence from the set that converges to it), uses (part of ) first countability. 
Also, the direction $f$ continuous implies $f$ preserves all sequence limits, holds for all spaces $X,Y$ and $f$ between them. Only the sufficiency needs some assumption on $X$, but no assumptions on $Y$ (or $f[X]$).
