I'm reading Komornik's Topology, Calculus and Approximation. Here:
I don't quite understand the proof. We're trying to prove that $a\in F$, right? I don't see how this proof implies that. I've been able to prove it using something else but I don't understand the proof via this lemma.
$F$ is not closed iff $X \setminus F$ is not open. By definition (i.e. iff), this means that there is a point $a \in X \setminus F$ such that there is no open set containing $a$ contained in $X \setminus F$. Equivalently, there is a point $a \in X \setminus F$ such that every open set containing $a$ is not contained in $X \setminus F$, i.e. every open set containing $a$ meets $X \setminus (X \setminus F) = F$. You can replace open set here with open ball.
Now apply Lemma 1.3 with $D = F$. The statement every open set containing $a \in X \setminus F$ meets $D = F$ is equivalent to the existence of a sequence $(x_n)$ in $D = F$ converging to $a \in X \setminus F$.
If $P$ is the statement $F$ is closed and $Q$ is the statement every convergent sequence contained in $F$ converges in $F$ then this argument establishes $\neg P \iff \neg Q$, since we just showed $F$ is not closed iff there is a point $a \in X \setminus F$ and a sequence $(x_n)$ in $F$ converging to $a$. But $\neg P \iff \neg Q$ is equivalent to $P \iff Q$.