Summary: Let $(X,\tau)$ be a topological space and $A\subseteq X$. Is the first countability of $X$ needed for proving that any point $x$ is in the interior of $A$ iff every sequence in $X$ converging to $x$ is eventually in $A$?
$\newcommand{\nat}{\mathbb{N}}$ I've proved that when $X$ is first-countable, any point is in the closure of $A$ iff there is a sequence in $A$ that converges to $x$, that is, $$ \forall x \in X:(x \in \overline{A} \iff \exists(x_i)_{i\in\nat} \subseteq A: \lim_{i \rightarrow \infty}x_i = x) $$ where I used the first accountability in the construction of such a sequence ($\Rightarrow$ direction).
But I spent almost a day to figure out whether I even need first accountability in proving that any point $x$ is in the interior of $A$ iff every sequence in $X$ converging to $x$ is eventually in $A$, or $$ \forall x \in X:(x \in A^{\circ} \iff \forall(x_i)_{i\in\nat} \subseteq X: (\lim_{i \rightarrow \infty}x_i = x \Rightarrow \exists j \in \nat:\forall k \geq j:x_k \in A)) $$
In my attempt in proving it, I didn't use the first accountability in both directions. It was easy to show the sufficiency ($\Rightarrow$ direction) and it does not require the first countability. When showing the other direction, I tried to show the contrapositive of the proposition using proof by contradiction. $$ \forall x \in X:(\exists(x_i)_{i\in\nat} \subseteq X: (\lim_{i \rightarrow \infty}x_i = x \land \forall j \in \nat:\exists k \geq j:x_k \not\in A) \Rightarrow x \not\in A^{\circ}) $$ Unlike the closure case, this form does not require the construction of the sequence because we can just assume it exists. If we suppose $x \in A^\circ$, we end up with a contradiction due to $\forall j \in \nat:\exists k \geq j:x_k \not\in A$.
Solved: I mistook the direction of the contrapositive of the necessity ($\Leftarrow$), which means I have to actually construct a sequence. $$ \forall x \in X:(x \not\in A^{\circ} \Rightarrow \exists(x_i)_{i\in\nat} \subseteq X: (\lim_{i \rightarrow \infty}x_i = x \land \forall j \in \nat:\exists k \geq j:x_k \not\in A)) $$
Here is an example showing that it can fail if $X$ is not first countable.
Let $X$ be an uncountable set, and let $p\in X$. Let
$$\tau=\{U\subseteq X:p\notin U\text{ or }X\setminus U\text{ is countable}\}\,;$$
$\tau$ is a topology on $X$ making $X$ a Lindelöf Hausdorff space. This topology is not first countable at $p$. (This is a good exercise.) Let $A=\{p\}$. Then $A$ is not open in $X$, but a sequence $\langle x_n:n\in\Bbb N\rangle$ in $X$ converges to $p$ if and only if there is an $m\in\Bbb N$ such that $x_n=p$ for all $n\ge m$, i.e., if and only if the sequence is eventually in $A$.