Showing that if a sub-subsequence converges then the sequence also converges 
Let $x \in X$ and $x_n$ be a sequence in $X$. If every subsequence of $x_n$ has a subsequence that converges to $x$. Show that $x_n$ must also converge to $x$.

I'm trying to get a contradiction here by supposing the opposite. So suppose that $x_n$ does not converge to $a$. This means that there exists $\varepsilon >0$ such that $d(x,x_n)\ge\varepsilon$  for some $n \ge K \in \mathbb{N}$. I'm not sure how I should continue from here? What definitions can I use to proceed?
 A: The negation of convergence of $x_n \to x$ is:
$$\exists \varepsilon>0: \forall N \exists n > N: d(x_n, x) \ge \varepsilon\tag{1}$$
Fix this $\varepsilon>0$ for the remainder.
Now proceed by recursion: Let $n_1$ be the $n$ given (by $(1)$) for the choice $N=1$, so that we know $d(x_{n_1}, x) \ge \varepsilon$.
Now having defined $n_1 < \ldots < n_k$ in $\Bbb N$ so that $x_{n_k}$ satisfies $d(x_{n_i}, x) \ge \varepsilon$, for all $i \le k$, pick $N=n_k$ in $(1)$ and we have some $n$, which we call $n_{k+1} > n_k$ and which also obeys $d(x_{n_{k+1}}, x) \ge \varepsilon$.
So from the negation of $x_n \to x$ we have constructed a subsequence $(x_{n_k})_k$ of $(x_n)_n$ with all terms at least $\varepsilon$ away from $x$.
It follows that no subsequence of $(x_{n_k})_k$ can converge to $x$, as the open ball $B(x,\varepsilon)$ contains no points at all of that subsequence. This contradicts the assumption on $(x_n)_n$ so $x_n \to x$ must hold.
A: You are almost done. For any $N$ there is an $n>N$ such that $|x_n-a| > \epsilon.$ Now, take the subsequence $(x_n)$ with $n$ as in the previous sentene.
