You are using the following fact:
Characterization of openness in first countable spaces:
In a first countable space, a set $U$ is open if and only if every sequence that converges to a point in $U$ is eventually contained in $U$.
You appeal to this in the last sentence of your proof; so this is where you are using first countability.
As for why the characterization of openness above relies on first countability, let's look at it's proof.
One way to prove the reverse implication of the Characterization of openness theorem (which is what you use in your proof) is to use the Characterization of closedness in first countable spaces:
Characterization of closedness in first countable spaces:
If $X$ is first countable and $A\subset X$, then $x\in\overline A$ if and only if there is a sequence $(x_n)$ contained in $A$ which converges to $x$.
The proof of the forward implication of Characterization of closedness theorem uses first countability in an essential way: If $x\in \overline A$, pick a countable decreasing nhood base $\{U_n\}$ at $x$. Then choosing $x_n\in U_n\cap A$ provides the required sequence (note that since $x\in \overline A$ the sets $U_n\cap A$ are nonempty).
Now, back to the prove of the reverse implication of the Characterization of openness theorem. We proceed with the proof of the contrapositive:
Let $A$ be a subset of $X$ that is not open. Then there is a point $x$ in $A$ that is also in the closure of $A^C$. From the Characterization of closedness theorem, there is a sequence in $A^C$ that converges to $x$. Such a sequence cannot be eventually contained in $A$. (One could also argue directly here: every nhood of $x$ contains points of $A^C$. So, select a decreasing countable nhood base at $x$ from which to find a sequence $(x_n)\subset A^C$ that converges to $x$.)
Note that Brian M. Scott's answer shows that first countability is necessary for the reverse implication in the Characterization of openness theorem.