First-countable space I want to show that if $X$ is first-countable and $x_n \to x$ implies $f(x_n) \to f(x)$ then $f: X \to Y$ is continuous.
That is for a neighborhood $V$ of $f(x)$ there is a neighborhood $U$ of $x$ such that $f(U) \subset V$.
So let $V$ be a neighborhood of $f(x)$. How to proceed from here?
 A: Lemma: Let $X$ be first countable. A set $U\subset X$ is open  if and only if whenever $x_n\rightarrow x\in U$, then $(x_n)$ is eventually in $U$ (that is, there is a positive integer $n$ so that $x_m\in U$ for each $m\ge  n$).
Proof: As the forward implication is trivial, we only provide the proof for the reverse implication. Towards this end, suppose $U$ is not open. Let $x$ be a point in $U$ such that no nhood of $x$ is contained in $U$. Let $\{ U_i\mid i\in\Bbb N\}$ be a countable nhood base at $x$ such that $U_1\supset U_2\supset U_3\supset\cdots$ (replace $U_n$ with $\cap_{k=1}^n U_n$ if necessary). For each $i$, choose $x_i\in U_i$ such that $x_i\notin U$. Then $(x_i)$ converges to $x\in U$ but is not eventually in $U$.

Now to show that your statement is true:
Let $V$ be open in $Y$. Set $U=f^{-1}(V)$ and suppose $x\in U$. 
Let $(x_n)$ be a sequence in $X$ converging to $x$.  Then $(f(x_n))$ converges to $f(x)\in V$.  Since $V$ is open the  the sequence $(f(x_n))$ is eventually in $V$ by the Lemma. From this, it follows that the sequence $(x_n)$ is eventually in $U$. Thus, by the Lemma again, $U$ is open in $X$.  
As $V$ was an arbitrary open set in $Y$, it follows that $f$ is continuous.
