Here is the setup for the problem I'm trying to solve.
Let $a_n$ be a bounded sequence of real numbers and $A$ the set of subsequential limits of $A$. Prove that $A$ is compact.
In $\mathbb{R}$, I can show that $A$ is closed and bounded, which will imply compactness. I am not completely sure how to show this, but here is my attempt.
Bounded: $\{a_n\}$ is bounded, so there exists $M > 0$ such that $|a_n| < M$ for all $n$. Let $a \in A$. So for some subsequence $\{a_{n_k}\}$, we have $(a_{n_k}\} \to a$. So there exists $N$ such that for every $n_k > N$, $|a_{n_k} - a| < 1$. Then, for $n_k > N$, $$ |a| = |a - a_{n_k} + a_{n_k}| \leq |a - a_{n_k}| + |a_{n_k}| < 1 + M. $$ This is an arbitrary element of $A$, so it's true for every element of $E$. So $1 + M$ is a bound for $A$.
Closed: Let $p$ be a limit point of $A$ and $r > 0$. I need to show that $p \in A$, meaning that $p$ itself is a subsequential limit of $a_n$. For $n > 0$, the open ball $B_{\frac{1}{n}} (p)$ contains some points of $A$ which are not equal to $p$. Pick one and call it $p_n$. Then $p_n \to p$.
I'm stuck at this point. I have a sequence in $A$ that converges to $p$, but I need a subsequence of $\{a_n\}$ that converges to $p$. Every $p_n$ is a subsequential limit itself, so I'm not sure if I can pick elements of individual subsequences arbitrarily close to $p_n$ so to maintain the limit $p$.
