Convergence of sequence and interior points For a subset $A \subseteq X$, consider the statement, "$x$ is an interior point of $A$ iff for every sequence $(x_m)$ in $X$ converging to $x$ there exists $n \in \mathbb{N}$ such that for all $m > n$ the term $x_m$ lies in $A$".
Now I know this theorem is true when $X$ is a metric space. But is it true for general topological spaces?
I am able to prove first part of theorem(just using definition of interior). But what about if If I start with for all sequence above statement true and then proceed?
 A: The converse can be proved by showing that for $x\notin\text{int}\left(A\right)$ a sequence
$\left(x_{n}\right)$ can be constructed that converges to $x$ with
$x_{n}\notin A$ for each $n$. 
This can be done if we are dealing
with a first-countable space. 
Let $\mathcal{V}=\left\{ V_{n}\mid n\in\mathbb{N}\right\} $
be a countable neighborhood basis of $x$ with $V_{1}\supset V_{2}\supset V_{3}\supset\cdots$. 
Then $V_{n}\cap A^{c}\neq\emptyset$ and if $x_{n}\in V_{n}\cap A^{c}$
then $\left(x_{n}\right)$ is a sequence outside $A$ that converges to $x$.

edit:
The converse is not true in general as is shown below.
Let $\left(\mathbb{R},\tau\right)$ be a topological space where $U\in\tau$
iff $U=\emptyset$ or $U^{c}$ is countable, and let $\left(x_{n}\right)$ be a sequence converging to $x$. 
Suppose $\left(x_{n}\right)$ has a subsequence $\left(x_{n_{k}}\right)$
with $x_{n_{k}}\neq x$ for each $k$. Then $U=\left\{ x_{n_{k}}\mid k\in\mathbb{N}\right\} ^{c}$
is an open set that contains $x$ and no $m$ can be found with $n>m\Rightarrow x_{n}\in U$,
contradicting the convergence. So we conclude that $\left(x_{n}\right)$
has no subsequence like that, which means that $x_{n}=x$ for $n$
large enough. Setting $A=\left\{ x\right\} $ proved is now that for
any subsequence $\left(x_{n}\right)$ convergent to $x$ some $m$
exists with $n>m\Rightarrow x_{n}\in A$. However, $x$ is not an
interior point of $A=\left\{ x\right\} $
