If all proper subsequences converge to same limit then the sequence converges. Let $\{X_n\}_n$  be a bounded sequence. Its convergent proper subsequences converge to the same limit $\ell$.
I want to prove that $\{X_n\}_n$ converges to $\ell$.
Notice that proper subsequences are all the sequences except for the sequence itself.
Is it enough to say that $\{X_{2n}\}$ and $\{X_{2n+1}\}$ are convergent to $l$ then $\{X_n\}$ is convergent to $\ell$?
 A: Proof by contradiction
Suppose that $\{X_n\}$ does not converge to $\ell$. Then, there is $\varepsilon_0>0$ such that $$\forall N\in\mathbb N,\exists n=n(N) : n>N~~~and ~~~ |X_n -\ell|>\varepsilon_0 $$
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 converging 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
$$0=\lim_{p\to\infty } |X_{ n_{k_p} }-\ell|>\varepsilon_0>0~~~\text{which is a CONTRADICTION}$$

Note that
$$\forall p,~~|X_{ n_{k_p} }-\ell|>\varepsilon_0$$
Since
$$\forall k,~~|X_{ n_{k}} -\ell|>\varepsilon_0$$

