Subsubsequence converges $\implies$ sequence converges 
Prove that if $\left\{ x_n \right\}$ is an infinite sequence of real numbers, $x \in \mathbb{R}$, and every subsequence $\left\{ x_{n_k} \right\}$ has a subsequence $\left\{ x_{n_{k_j}} \right\}$ with $x_{n_{k_j}} \rightarrow x$, then $x_n \rightarrow x$.

I know that if every subsequence of a sequence converges to the same number, then the sequence converges to that same number. But I don't know if the same can be applied to subsubsequences. So for this problem, can I safely state that because $x_{n_{k_j}} \rightarrow x$, it is also true that $x_{n_k} \rightarrow x$? If this is true, then does that mean every subsequence $\left\{ x_{n_k} \right\}$ also converges to $x$?
 A: HINT: Suppose that $x_n\not\to x$, and show that $\langle x_n:n\in\Bbb N\rangle$ has a subsequence that is bounded away from $x$.
A: If $x_n$ doesn't converge to $x$, no tail of it converges. But then we can construct a subsequence of elements that are at least $\varepsilon$ from $x$, but that subsequence has to have a subsequence that converges to $x$. By contradiction $x_n$ converges to $x$.
A: Suppose that $\{X_n\}$ does not converge to $x$. Then, there is $\varepsilon_0>0$ such that $$\forall N\in\mathbb N,\exists \hspace{.2cm}n=n(N) : n>N~~~and ~~~ |X_n -x|>\varepsilon_0 $$
For $N_1=1$ there exists $n_1$ such that 
$$n_1>N_1 ~~~and ~~~ |X_{n_1} -x|>\varepsilon_0 $$
Taking successively $N_{k+1}> \max\{N_k, n_k,k+1\}$ there exists $n_{k+1>N_{k+1}}$ such that,
$$ |X_{ n_{k+1}} -x|>\varepsilon_0 $$
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, $$\forall k,~~ |X_{ n_{k}} -x|>\varepsilon_0  \qquad  \text{and}~~~\{X_{ n_{k}} \}~~~\text{is bounded} $$

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

