It can be proven that $$A_1, A_2 \subseteq \mathbb{R^n}, \space closed\space sets \implies A_1 \cup A_2 \space is \space closed$$
Using the definition of a closed set in vector spaces.
I have a proof here that considers the subsequences $(x_{k_i})$ of some sequence $(x_k) \in A_1\space or \space A_2$ and displays the convergence of these subsequences, i.e. that $(x_{k_i})\rightarrow a \in A_1 \space or \space A_2$.
I don't understand why one needs to consider the subsequences and not just the main sequence $(x_k)$. The proof talks about $(x_k)$ being infinite, is this the reason?
Let $(x_n)_{n\in\mathbb N}\subset A_1\cup A_2$ be a convergent sequence and denote the limit by $\ell$. Either $A_1$ or $A_2$ contains infinitely many terms of the sequence. Suppose without loss of generality that it's $A_1$ and denote $x_{n_k}$ those terms of $(x_n)_{n\in\mathbb N}$ that are in $A_1$. It's a subsequence of $(x_n)_{n\in\mathbb N}$ and since $(x_n)_{n\in\mathbb N}$ converges to $\ell$, $(x_{n_k})_{k\in\mathbb N}$ also converges to $\ell$. Since $A_1$ is close, $\ell\in A_1\subset A_1\cup A_2$. Therefore $A_1\cup A_2$ is close.