$X$ first-countable, $A \subseteq X$, $x\in X$, then $x \in \text{Int }A\Leftrightarrow$ every sequence in $X$ converging to $x$ is eventually in $A.$ I want my proof (below) for the following proposition verified.
Let $X$ be a first-countable topological space, let $A$ be a subset of $X$ and let $x\in X$. Then $x \in \operatorname{Int}A$   if and only if every sequence in X converging to $x$ is eventually in $A$.
Some definitions:


*

*$\operatorname{Int}A$ is the union of all open sets contained in $A$.

*"sequence is eventually in $A$" means that all but finitely many of the terms in the sequence are contained in $A$.
My attempt at the proof:
Suppose $x\in \operatorname{Int}A$
  and there exists a sequence $(x_i)$
  in $X$
  which converges to $x$
  and there are infinitely many $x_i$
  in $X-A$.
 Let $U$
  be a neighbourhood of $x$
  contained in $A$.
 Then there exists a positive integer $N$
  such that $x_i\in U$
  for all $i\geq N$.
 But then $U$
  is not contained in $A$,
 this is a contradiction. Therefore $x$
  can not be in $\operatorname{Int}A$. 
Conversely, suppose that $x\notin \operatorname{Int}$
  and let $\mathcal{B}_x= \{ U_i\} _{i=1}^\infty$
  be a countable neighbourhood basis at $x$.
  It is clear that each $U_i$
  is not contained in $A$.
 For each $n=1,2,\ldots$,
 choose a point $x_n\in U_1\cap\cdots\cap U_n$
  and consider the sequence $(x_n)$.
 Then $x_n\rightarrow x$ and this sequence is not eventually in $A$, in fact, none of the terms in sequence are in $A$. To see that  $x_n\rightarrow x$, let $U$
  be a neighbourhood of $X$,
 then there exists some positive integer $k$
  such that $U_k$
  is contained in $U$,
 and so $x_i\in U$
  for all $i\geq k$.
 A: You’ve made the first part much more difficult than necessary by bringing in $U$ and by arguing by contradiction; here’s a much simpler direct approach.

Suppose that $x\in\operatorname{int}A$, and $\sigma=\langle x_n:n\in\Bbb N\rangle$ is a sequence in $X$ converging to $x$. The interior of $A$ is an open nbhd of $x$, so there is an $m\in\Bbb N$ such that $x_n\in\operatorname{int}A$ for each $n\ge m$. And $\operatorname{int}A\subseteq A$, so $x_n\in A$ for each $n\ge m$, i.e., $\sigma$ is eventually in $A$.

The second part is wrong as stated: you’ve said nothing to ensure that the points $x_n$ are not in $A$. What you wanted, I suspect, is something like this:

Suppose that $x\in X\setminus\operatorname{int}A$, and let $\{U_n:n\in\Bbb N\}$ be a local base at $x$. For each $n\in\Bbb N$ let $B_n=\bigcap_{k\le n}U_k$, and let $\mathscr{B}=\{B_n:n\in\Bbb N\}$; $\mathscr{B}$ is then a nested local base at $x$. Since $x\notin\operatorname{int}A$, we know that $B_n\setminus A\ne\varnothing$ for each $n\in\Bbb N$, so for each $n\in\Bbb N$ we can pick a point $x_n\in B_n\setminus A$. If $V$ is any open nbhd of $x$, there is an $m\in\Bbb N$ such that $B_m\subseteq V$, and clearly $x_n\in V$ for each $n\ge m$, so $\langle x_n:n\in\Bbb N\rangle$ converges to $x$. By construction, however, the sequence is not eventually in $A$; indeed, it is never in $A$.

