Is the following true?
Let $x_n$ be a sequence with the following property: Every subsequence of $x_n$ has a
further subsequence which converges to $x$. Then the sequence $x_n$ converges to $x$.
I guess that it is true but I am not sure how to prove this.
True. If not, there exists an $\epsilon > 0$, such that for all $k$, there exists an $n_k > k$ satisfying $|x_{n_k}−x| \ge \epsilon$ since if there is some $k$ which doesn't have such $n_k$, then we can take it as $N$, so $x_n$ converges to $x$. The subsequence $x_{n_{k}}$ does not have any subsequence converging to $x$.
This is much easier to prove than the other answers here might suggest, if you remember the handy little fact (given, for instance, in Baby Rudin but which aught to be more well-known) that a sequence $x_n$ does not converge to $x$ if, and only if, there exists $\epsilon>0$ such that $|{x_n-x}| \geq \epsilon$ for infinitely many $n$.
On the one hand, if $x_n \to x$ then this fact gives that any subsequence $(x_{n_i}) \subset (x_i)$ can also have finitely many terms a distance $\epsilon$ from $x$, and therefore also converges to $x$. In particular, if $x_n$ converges to $x$ then every subsequence has a subsequence converging to $x$, namely itself
On the other hand, if $x_n \not\to x$ then there exist infinitely many $n$ such that $|x_n-x| \geq \epsilon$ and the subsequence of these terms has no subsequence converging to $x$ as any subsequence has all terms a distance of at least $\epsilon$ away from $x$
Note, however, that it is necessary that the sub-subsequnces converge to the same limit: for instance, any subsequence of the alternating sequence $0,1,0,1,0,1,\ldots$ has a convergent subsequence by the pigeonhole principle, but such sub-subsequences can converge to either 0 or 1
proof by Contradiction
Suppose that $\{X_n\}$ does not converge to $\ell$. Then, there is $\varepsilon_0>0$ such that $$\forall N\in\mathbb N,\exists \hspace{.2cm}n=n(N) : n>N~~~and ~~~ |X_n -\ell|>\varepsilon_0 $$
For $N_1=1$ there exists $n_1$ such that $$n_1>N_1 ~~~and ~~~ |X_{n_1} -\ell|>\varepsilon_0 $$ Taking successively $N_{k+1}> \max\{N_k, n_k,k+1\}$ there exists $n_{k+1>N_{k+1}}$ such that,
$$ |X_{ n_{k+1}} -\ell|>\varepsilon_0 $$
It is easy to see that, $\{X_{ n_k}\}_k$ is a subsequence of $\{X_{ n}\}_n$ since $$ n_k< n_{k+1} \quad i.e ~~\text{the map }~~k\mapsto n_k~~~\text{Is one-to-one}$$
However, $$\forall k,~~ |X_{ n_{k}} -\ell|>\varepsilon_0 \qquad \text{and}~~~\{X_{ n_{k}} \}~~~\text{is bounded} $$
Therefore By Bolzano-Weierstrass Theorem's there exists $\{X_{ n_{k_p} }\}_p$ subsequence of $\{X_{ n_{k} }\}_k$ which converges to some limit $\ell_1 $ but $\{X_{ n_{k_p} }\}_p\to \ell_1$ is also a congering subsequence of $\{X_n\}_n$
By assumption, $\ell=\ell_1$ that is together with the fact $\{X_{ n_{k_p} }\}_p$ is a subsequence of $\{X_{ n_{k} }\}_k$ we have
$$0=\lim_{p\to\infty } |X_{ n_{k_p} }-\ell|>\varepsilon_0>0~~~\text{CONTRADICTION}$$
Note that $$\forall p,~~|X_{ n_{k_p} }-\ell|>\varepsilon_0$$ Since $$\forall k,~~|X_{ n_{k}} -\ell|>\varepsilon_0$$
Direct proof(Update) Since $(|x_n-\ell|)_n$ is bounded, $M=\limsup_{n\to\infty} |x_n-\ell|$ exists and thus by definition of the $\limsup$, there exists a subsequnce $(x_{n_k})_k$ such that $$\lim_{k\to \infty} |x_{n_k}-\ell|=M= \limsup_{n\to\infty} |x_n-\ell| $$
By assumption there exists a further subsequence $(x_{n_{k_p}})_p$ converging to $\ell$. Hence we have $$0= \lim_{p\to \infty} |x_{n_{k_p}}-\ell|= \lim_{k\to \infty} |x_{n_k}-\ell|=M =\limsup_{n\to\infty} |x_n-\ell|.$$ Consequently, $$0\leq \liminf_{n\to\infty} |x_n-\ell|\leq \limsup_{n\to\infty} |x_n-\ell|=0$$
that is $$ \liminf_{n\to\infty} |x_n-\ell|= \limsup_{n\to\infty} |x_n-\ell|=0$$
that is $\lim_{n\to\infty} |x_n-\ell|=0$ or $x_n\to x$ as $n\to \infty$.
Assume $(x_n)$ doesn't converge to $x$. In case $1, (x_n)$ isn't bounded, thus has at least one sub-sequence going to either $\infty$ or $-\infty.$ This subsequence clearly doesn't have any subsequence converging to $x$ (thus a contradiction).
In case $2, (x_n)$ is bounded. If $(x_n)$ converges to a finite real limit different from $x$ then you get an immediate contradiction to the property. Otherwise $(x_n)$ is divergent and bounded thus has $\liminf$ and $\limsup.$ These two converge, and have a sub-sequence which converges to $x$, thus must converge to $x$ themselves (otherwise you get a similar contradiction). And this means that : $$\liminf=\limsup=0 \implies \lim(x_n)=x$$
