Suppose $(X,\tau)$ is a topological space and $A\subset X$. Let $\overline{A}$ denote the closure of $A$ in $X$. Suppose $x\in A$. Then there exists a net $\langle x_\delta\rangle_{\delta\in\Delta}$ such that $x_\delta\to x$.
A standard proof goes as follows.
Let $\Delta=\{\text{nbhds of } x\}$ and $U\le V$ in $\Delta$ if and only if $V\subset U$. For each $U\in\Delta$, choose $x_U\in U\cap E$. Then $x_U\to x$.
The definition of neighbourhoods (nbhds) of a point $x$ here is any subset of $X$ which contains $x$ as an interior point.
Now, assume in addition that $X$ is first countable. Can one use the net defined above to show that there is a sequence $\langle x_n\rangle_{n\in\mathbb{N}}$ which converges to $x$?
In a metric space, it is very easy to extract such sequence by considering the nested balls $\{B(x,\frac{1}{n})\}_{n\in\mathbb{N}}$. In general first-countable spaces, one looses the metric, but still has a countable local basis by definition. But I don't see how to exploit that to get a desired sequence.
Okay, assume that $X$ is first countable and let $y\in \overline{A}$ for some $A\subseteq X$. Let $(U_n)_{n\in \mathbb{N}}$ be a local neighbourhood base for $y$.
Since $y\in \overline{A}$, there exists $x_n\in A\cap (\cap_{i=1}^n U_i)$ for every $n$. Let's show that $x_n\to y$. Indeed, let $U$ be some open neighbourhood of $y$. Then, $U$ contains $U_m$ for some $m$. However, $x_n\in U_m$ for all $n\geq m$ and so, we see that $x_n\in U$ for every $n\geq m$. Since $U$ was arbitrary, we get that $x_n\to y$.