Let $X$ be a second countable topological space. Then $A \subset X$ is a compact subset of $X$ if and only if every sequence $\{x_{n}\}_{n \in \mathbb{N}} \subset A$ has a convergent sub-sequence with a limit point $x \in A$.
I can prove it when $X$ is a metrizable space, but I'm having trouble when $X$ is only second countable.
I'll appreciate any help. Thanks!
Suppose $A$ is compact. Suppose $(x_n)$ is a sequence in $A$. We can assume WLOG that $S=\{x_n \mid n \in \Bbb N\}$ is infinite, or we'd have a constant subsequence etc. As $A$ is compact, the infinite set $S$ has a condensation point $p \in A$, i.e. a point such that every neighbourhood of $p$ contains infinitely many points of $S$.
(The proof is simple: if no such point exists we get a quick contradiction with compactness: every point $a \in A$ would then have a neighbourhood $O(a)$ with $O(a) \cap S$ finite and a finite subcover gives a contradiction..)
As this $p$ has a decreasing neighbourhood base (as $X$ is in particular first countable), this allows us to define a subsequence of $x_n$ that converges to $p$ by a straightforward recursion. (Note that we only used first countability here.)
For the reverse: if $A$ obeys the sequential compactness condition, it is easily seen to be countably compact in the sense that every countable cover has a finite subcover (because the covering version is equivalent to the fact that every countably infinite set has a condensation point (or "strong limit point"), see e.g. part of my answer here ) and we can arrange an infinite subset in a bijective sequence so that a limit of a subsequence will be a condensation point of that set. As $X$ is second countable, all subsets of it, so $A$ too, is second countable and hence Lindelöf. And a Lindelöf countably compact subset is compact. QED.
Note that all of this is separation axiom free, BTW.