Convergence of sequence of real numbers and (sub)subsequence Let $(x_n)_{n\in \mathbb{N}}$ be a sequence of real numbers. A well known result is:

The sequence $(x_n)_{n\in \mathbb{N}}$ converges to $a\in \mathbb{R}$ if and only if every subsequence $(x_{n_k})_{k\in \mathbb{N}}$ converges to $a\in \mathbb{R}$.

Now, I'm working on an exercise which essentialy asks to prove the following:

The sequence $(x_n)_{n\in \mathbb{N}}$ converges to $a\in \mathbb{R}$ if and only if for every subsequence $(x_{n_k})_{k\in \mathbb{N}}$ there is a (sub)subsequence $(x_{n_{k_j}})_{j\in \mathbb{N}}$ which converges to $a$.

Now, one direction is trivial. If in fact $(x_n)_{n\in \mathbb{N}}$ converges to $a$, then by the result I've mentioned, every subsequence of $(x_{n})_{n\in \mathbb{N}}$ converges to $a$. Let then $(x_{n_k})_{k\in \mathbb{N}}$ be an arbitrary subsequence. In that case, we know that it is a subsequence of itself, and hence we've found a (sub)subsequence which converges to $a$.
The other way around seems quite complicated for me. The reason is as follows: I must assume that every subsequence $(x_{n_k})_{k\in \mathbb{N}}$ has a (sub)subsequence which converges to $a$ and the proof that the original one converges to $a$.
My problem is exactly that "a". If it had all (sub)subsequences converging to $a$, then by the result i've mentionted, the subsequence $(x_{n_k})_{k\in \mathbb{N}}$ would converge to $a$. In that case, we would have every subsequence converging to $a$ and the result would follow.
My idea to prove the claim was then:
First remember that a subsequence is the restriction of the original one $x: \mathbb{N}\to \mathbb{R}$ to an infinite subset $\mathbb{N}'\subset \mathbb{N}$. In that case a (sub)subsequence is the restriction of $\tilde{x} = x|\mathbb{N}'$ to an infinite subset $\mathbb{N}''\subset \mathbb{N}$. But since $\tilde{x}$ is the restriction of $x$ to $\mathbb{N}'$ and since $\mathbb{N}''\subset \mathbb{N}'$, the restriction of $\tilde{x}$ to $\mathbb{N}''$ is itself the restriction of $x$ to $\mathbb{N}''$.
In other words, every (sub)subsequence can be viewed as a subsequence of the original one. Considering this, by hypothesis every subsequence has a (sub)subsequence which converges to $a$. But by this argument, this means that we can replace every subsequence with a convergent one. And then we can use the result I stated.
Is my idea correct? Is this really the way to prove the claim? My only doubt is that I need every subsequence converging to $a$ to use the result. In this case I can't see how I can argue that this implies every subsequence converges to $a$.
How can I finish this proof?
 A: By contrapositive: Suppose that $(x_n)\not\rightarrow a.$ This means that $\exists\delta>0$ such that $\forall N\in\mathbb N$ there is some natural $n>N$ such that $|x_n-a|\geqslant\delta.$ Now construct a subsequence $(x_{n_k})$ of $(x_n)$ as follows.  Let $n_0>0$ be such that $|n_0-a|\geqslant\delta.$ Having chosen $n_0,\ldots,n_k,$ let $n_{k+1}$ be such that $n_{k+1}>n_k$ and $|x_{n_{k+1}}-a|\geqslant\delta.$ Clearly, the subsequence $(x_{n_k})$ does not have a subsequence which converges to $a.$ Consequently, if $(x_n)\not\rightarrow a,$ you will always be able to find a subsequence $(x_{n_k})$ of $(x_n)$ that is far from $a.$ Therefore, if every subsequence of $(x_n)$ has a subsequence which converges to $a$ then $(x_n)$ converges to $a.$ 
A: If you are allowed to appeal to point-set topology:
To prove the converse, note that since for every $(x_{n_{k}})$ there is some $(x_{n_{k_{j}}})$ $\to a$ by assumption, by definition $a$ is the limit point of every subsequence of $(x_{n})$, so $\{ x_{n} \} \cup \{ a \}$ is compact. Since $a$ is the unique limit point of $(x_{n})$, it follows that $a$ is the limit of $(x_{n})$.
