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 $x$.
Suppose, for example, that the space is $\Bbb R$, and $A=\left\{\frac1n:n\in\Bbb Z^+\right\}$; the only limit point of $A$ is $0$. If $n\ge 2$, $\left(\frac1{n+1},\frac1{n-1}\right)$ is an open interval around $\frac1n$ whose intersection with $A$ is just $\left\{\frac1n\right\}$, and $\left(\frac12,2\right)$ is an open interval around $1$ whose intersection with $A$ is just $\{1\}$, so each point of $A$ has an open nbhd that contains no other point of $A$. Morover, none of these nbhds contains the limit point $0$. The thing that makes $0$ a limit point of $A$ is that every open nbhd of $0$ contains points of $A$ different from $0$.
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.
A compact Hausdorff space need not be sequentially compact, there is an example here (near the end of the page), and a sequentially compact space need not be compact (e.g. the first uncountable ordinal $\omega_1$ in the order topology, which is first countable, hereditarily normal, sequentially compact, countably compact, but not compact).
Of course a compact space is trivially countably compact: if all open covers have finite subcovers, all countable ones certainly do.
A sequence $(x_n)$ (say of all distinct terms) in a countably compact space $X$ can be seen as an infinite set, and so it has an $\omega$-accumulation point (every neighbourhood of $p$ contains infinitely many terms of the sequence). See my answer here.
But this does not guarantee that there is a subsequence that converges to $p$ (see the proof in the link I gave), because there can be a lot of neighbourhoods of $p$, and we need every neighbourhood of $p$ to contain a tail of the subsequence.
If however $p$ has a countable local base $O_n$ (like in metric spaces, where we can use the balls around of $p$ of radius $\frac{1}{n}$), then we can find such a subsequence: pick $n_1$ with $x_{n_1} \in O_1$, $n_2 > n_1$ with $x_{n_2} \in O_1 \cap O_2$, $n_3 > n_2$ such that $x_{n_3} \in O_1 \cap O_2 \cap O_3$ and so on. This uses that all these (finite!) intersections are neighbourhoods of $p$ and thus contain infinitely many terms of the sequence. And the fact that these $O_n$ form a local base, ensures that the subsequence $x_{n_k}$ converges to $p$ as $k$ tends to infinity: if $O$ is open and contains $p$, for some $m$, $O_n \subset O$, and so all $x_{n_k}$ with $k \ge m$ are by construction in $O_m$ and hence in $O$.
So in first countable (countably) compact spaces we do have sequential compactness, and
many spaces are indeed first countable (like metrisable spaces), so most people's intuition is more tuned to that case. But in general we do not have that compact spaces are sequentially compact.
Best Answer
No, this is in fact equivalent (in ZF) to the statement that every infinite set has a countably infinite subset. Indeed, suppose $S$ is an infinite set with no countably infinite subset. Then any sequence in $S$ can take only finitely many values (otherwise its image would be a countably infinite subset of $S$), and so has a convergent subsequence with respect to any topology (just pick a subsequence that is constant). In particular, giving $S$ the discrete topology, $S$ is sequentially compact but not limit point compact.