We use the sequential definition to prove a set is closed. So no continuity or closure or anything related to the topology of the set is allowed.
Show $A = \{ x \in \ell^2: |x_n| \leq 1/n \}$ is closed.
We prove a weaker statement first. We show for a fixed $k$, the set $A_k = \{ x \in \ell^2: |x_k| \leq 1/k \}$ is closed and then later argue that $\cap A_k$ is closed.
So let $x^i \in A_k \subset \ell^2$, assume $x^i \to x \in \ell^2$. Required to show $x^*_k \in A_k.$
Since $x^i$ converges in $\ell^2$, $\exists N_1 \ni i > N_1 \implies \| x^i - x\|_2 < \epsilon.$
Now, $|x_k^i - x_k^*| \leq \|x^i - x\|_2 < \epsilon$ meaning $\lim_{i \to \infty} x_k^i = x_k^*$. Now for each $k$, we have $-1/k \leq x_k^i \leq 1/k$, so squeeze theorem says $-1/k \leq x_k^* \leq 1/k$ (notice this limit is independent of $k$). This shows that $x^*_k \in A_k$, meaning $A_k$ is closed for each $k$.