Def'n 1. is the general def'n of compactness in topology, whether or not the topology can be generated by a metric.
In a metric space, a set $X$ is compact iff every sequence in $X$ has a limit point that BELONGS to $X$ iff every infinite subset of $X$ has a limit point that belongs to $X.$
The n-dimensional generalization of the (1-dimensional) Bolzano-Weierstrass theorem is that a subset of $R^n$ is compact iff it is closed and bounded. (But this does not hold for all metric spaces.)
It is useful to know various equivalents to compactness, and to know that there are equivalents specific to metric spaces, just as it is useful to know various equivalents of "continuous function", some of which are specific to metric spaces.
Another term is "pre-compact" which I have seen only in the context of Hausdorff spaces : $X$ is pre-compact iff $\overline X$ is compact.
Yes, it is well-known that
Let $X$ be a sequential space. Then $X$ is countably compact iff $X$ is sequentially compact.
To be clear about definitions: $X$ is countably compact iff every countable open cover of $X$ has a finite subbcover. I'll use the convenient equivalence that $X$ is countably compact iff every infinite subset $A$ of $X$ has an $\omega$-accumulation point $p \in X$, i.e. every neighbourhood $U$ of $p$ has $U \cap A$ infinite. I wrote the proof of that equivalence (also free of separation axiom assumptions) here, in case this was unknown to you.
And $X$ is sequentially compact iff every sequence $(x_n)_n$ in $X$ has a convergent subsequence.
$X$ is sequential if for all subsets $A$ of $X$: $A$ sequentially closed iff $A$ is closed.
Proof (following the standard reference Engelking, General Topology 2nd ed. Theorem 3.10.31 (due to Franklin (1965), generalising this from metric spaces where it had been shown by Hausdorff in 1914). I'll be extra pedantic in my version of its proof to make sure I make no hidden separation axiom assumptions (Engelking includes Hausdorff in his definition of countable compactness and sequential compactness, so there are subtle instances in his proofs where he might use them, which I want to avoid for maximal generality)
If $X$ is sequentially compact, let $A$ be an infinite subset of $X$. Let $(a_n)_n$ be sequence of points from $A$ so that $a_n \neq a_m$ whenever $n \neq m$ (i.e. find an injection $\Bbb N \to A$). By sequential compactness, there is some $p \in X$ and some subsequence $(a_{n_k})_k$ of $(a_n)_n$ such that $a_{n_k} \to p$ as $k \to \infty$. Then if $U$ is any neighbourhood of $p$, there exists some $N$ so that for all $k \ge N$ we have $a_{n_k} \in U$. It follows that $\{a_{n_k} \mid k \ge N\} \subseteq U \cap A$ and so $U \cap A$ is infinite (by the injectivity). Hence $p$ is an $\omega$-limit point of $A$ and by the mentioned equivalence, $X$ is countably compact. This shows one implication.
Let $X$ be countably compact. Let $(x_n)_n$ be a sequence in $X$. We want to show it has a convergent subsequence.
We can assume WLOG that $n \neq m \to x_n \neq x_m$ (if you believe this read on, if you want my argument for it, reveal spoiler)
For $x \in X$ we can define $N_x=\{n \in \Bbb N\mid x_n = x\}$. If some $N_x$ is infinite, then $N_x$ defines a subsequence of $(x_n)_n$ that is constant with value $x$ and in any space this converges to $x$, and we'd be done. So assume all $N_x$ are empty (probably most of them are) or finite; picking one index from each non-empty one (yes I believe in AC) we get a subsequence as claimed, and as a subsequence of a subsequence is still a subsequence, we only have to consider that situation. We resume the proof under this injectivity assumption which is a handy technicality.
Define $A = \{x_n\mid n \in \Bbb N\}$ which is an infinite set. As $X$ is countably compact, $A$ has an $\omega$-accumulation point $p$. It's clear that the $A\setminus \{p\}$ is not closed, as $p$ is in its closure but not in the set. Because $X$ is sequential (finally we use it!), $A\setminus \{p\}$ is not sequentially closed, i.e. there is a sequence $(y_n)_n$ in $A \setminus \{p\}$ and some point $q \notin (A\setminus \{p\})$ so that $y_n \to q$.
By re-ordering this sequence we find a subsequence of $(x_n)_n$ converging to $q$. (details follow as before, skip if you see it already).
First note as all $y_n$ come from $A\setminus \{p\}$ and the sequence is injective, for every $n$ there is a unique $r(n) \in \Bbb N$ so that $y_n = x_{r(n)}$. Another folklore fact: if $h: \Bbb N \to \Bbb N$ is a bijection then $(y_{h(n)})_n \to q (n \to \infty)$ as well (any re-ordering of a convergent sequence has the same limit): let $U$ be any neighbourhood of $q$. There is an $N_1 \in \Bbb N$ so that $n > N_1$ implies $y_n \in U$. Then $h^{-1}[\{1,\ldots, N_1\}] \subseteq \{1,\ldots, N\}$ for some $N \in \Bbb N$, as $h^{-1}$ preserve finiteness. Then $h(n) > N$ implies $n > N_1$ so $q \in U$, and as $U$ was arbitrary, $y_{h(n)} \to q$ as $n \to \infty$. Then enumerate $r[\Bbb N]$ as $n_1< n_2< n_3 <n_4< \ldots$ and note that $x_{n_k} = (y_{r^{-1}(k)})$ converges to $q$ (as $k \to \infty$) as a re-ordering of $(y_n)_n$. So we have the desired convergent subsequence.
QED.
Best Answer
There’s no reason to choose a nbhd of $x_1$. When we say that $x$ is a limit point of $A$, we’re saying that every open nbhd of $x$ contains a point of $A$ different from $x$, so $x$ is the only point whose nbhds are of interest. It’s entirely possible that for each $n\in\Bbb N$ the point $x_n$ has an open nbhd that contains no other point of $A$ and does not contain the limit point $x$.
Now go back to the general situation. If $A$ is infinite, limit point compactness says that it must have a limit point $x$. This means that for each $\epsilon>0$, $B(x,\epsilon)\cap(A\setminus\{x\})\ne\varnothing$: each $\epsilon$ ball centred at $x$ contains a point of $A$ other than $x$. (Of course it’s possible that $x\notin A$, as in the example above, but we have to cover the possibility that $x$ is in $A$ as well.) In particular, for each $k\in\Bbb Z^+$ there some $x_{n_k}\in B\left(x,\frac1k\right)\cap(A\setminus\{x\})$. Then $\langle x_{n_k}:k\in\Bbb Z^+\rangle$ is a sequence in $A$ converging to $x$. The only problem is that it might not be a subsequence of the original sequence $\langle x_k:k\in\Bbb N\rangle$, because the indices $n_k$ might not be strictly increasing. To complete the proof, you must show that the sequence $\langle n_k:k\in\Bbb Z^+\rangle$ of indices has a strictly increasing subsequence.