Consider the following statement
$x_n \to x$ if and only if every subsequence of $x_n$ has a subsequence that converges to $x$.
$\implies$ is clear. A proof of the other direction is given here. It is a proof by contrapositive.
My question is: can it be proved directly? (I obviously tried but couldn't do it)
As Asaf Karagila noted in the comment, the referenced proof is rather indirect. Also note that these proof types are not formally defined so it is not clear what will you accept as direct proof and what not.
The core argument of given proof is as follows: a sequence $S$ which doesn't converge to a point $x$ has a subsequence which completely misses some neighborhood of $x$ and so none of its subsequences can converge to $x$.
I'll put down something which looks more like a direct proof. First some definitions. We'll say that a sequence $S$ is eventually in a set $U$ iff $U$ contains a final segment of $S$. And we'll say that $S$ is frequently in $U$ iff it contains a subsequence which is eventually in $U$.
Note that for each $S$ and $U$ exactly one of the following cases holds:
Also note that $S \to x \iff$ $S$ is eventually in any neighborhood of $x$.
Now see the following steps:
Is this a direct proof for you?
It is unclear whether you mean a metric or a topological space. I will assume (from the link) you mean metric.
I have proved this before in my analysis course I think, but not directly.
Suppose every convergent subsequence we can construct in this manner belong to the set $A$. Take the union of ${x \in x_{n_{k}} \in A}$ to obtain the set $B$ of all elements which belong to a subsequence of $x_n$ which converges to $x$. Let $X$ be the set of all elements in $x_n$. Clearly, $X \setminus B$ is finite, as if we suppose not then we can find another subsequence of $x_n$ with elements in $X \setminus B$ which has a subsequence which converges to $x$. This subsequence belongs to $A$ so its elements belong to $B$ so its elements don't belong to $X \setminus B$, contradiction. It is now sufficient to show that the subsequence of $x_n$ containing all elements in $B$ converges to $x$ (as there are finitely more elements of $x_n$ not in this subsequence). Call it $x_{n_{k}}$. It satisifies all the conditions that $x_n$ satisfied, so we can repeat this process, to get $x_{n_{k2}}$, etc.
This is the subtle part. If, repeating this process, we never get a subsequence of $x_n$, say $x_{n_{kr}}$, with $x_{n_{kr}} = x_{n_{k(r-1)}}$ (as sequences) then, ad infinitum, we get an infinite set of points in $X$ which must be members of $X \setminus B$. Contradiction. For this $r$, $x_{n_{kr}}$ is a sequence with the following properties: all of what $x_n$ had, and the fact that the subsequences of the subsequences which converge to $x$ cover $x_{n_{kr}}$ (interpreted intuitively, as with the sets presented before), $x_{n_{kr}}$. So we can find a finite subcover (ok, I'm assuming compact too). Hence, $ \forall \epsilon > 0 \exists N_1,...,N_k$ after which these subsequences converging to $x$ are within $\epsilon$ of $x$ (notation gets out of hand here, so just being wordy, you know what I mean). Then for $n \geq max(N_1,...,N_k)$, the nth element onwards of $x_{n_{kr}}$ is within $\epsilon$ of $x$. As there are finitely many elemens ot $x_n$ not in $x_{n_{kr}}$, the same holds true for $x_n$, i.e. it converges to $x$.
Note: I am out of time so if someone could help with the notation and explanation I'd be hugely grateful :).
