Convergence of a sequence by convergence of sub-subsequence Suppose that $\{p_n\}_{n \in \mathbb{N}}$ is a sequence in a metric space $X$. Assuming that every subsequence of $\{p_n\}_{n \in \mathbb{N}}$ has itself a subsequence that converges, say, to $p$, show that $\{p_n\}_{n \in \mathbb{N}} \to p$.  
Solution Attempt:
Take $ \mathbb{N} \supset K := \{n,n+1,n+2, \ldots\}$ for $n \in \mathbb{N}$ to be our subsequence. Now, there exists $J_1 \subset K \subset \mathbb{N}$ so that $\{p_j\}_{j \in J_1} \to p$. If $J_1 = K$, we're done, otherwise assume $J_1 \subset K$.
Thus, $K \setminus J_1$ is non-empty, take it to be our subsequence. Thus, there exists $J_2 \subset (K \setminus J_1)$ so that $\{p_j\}_{j \in J_2} \to p$. If $J_2 = (K \setminus J_1)$, we're done, otherwise assume $J_2 \subset (K \setminus J_1)$ so that the next iteration is non-empty.
Inductively let, $$K \setminus \bigcup_{i=1}^{k} J_i$$ be our next subsequence so that there exists $$J_{k+1} \subset \Big(K \setminus \bigcup_{i=1}^{k} J_i \Big)$$ so that $\{p_j\}_{j \in J_{k+1}} \to p$. It is possible that we are left with a finite set $B \subset K$ so that $B$ is never in $J_k$. In this case, set $m = \max\{B\}$, so that the original sequence converges above $m$. Thus, $\{p_n\}_{n \in \mathbb{N}} \to p$ as desired.  
Is this a correct proof? I proved the statement by contradiction rather easily, but I wanted to try my hand at a direct proof. Any comments are appreciated.
 A: This should work for real sequences, but not arbitrary metric spaces. I'm trying that out now. 
Suppose every subsequence has a convergent subsequence with limit $p$.
Let $\alpha = \limsup_\limits{n\to \infty} p_n$.
Then by the first answer here:
http://www.stat.uchicago.edu/~lekheng/courses/104s09/math104s09-hw5sol.pdf
there exists a subsequence $\{x_{n_k}\}$ that converges to $\alpha$ as $k \to \infty$.
Then that subsequence has a convergent subsequence that converges to $p$ by assumption. But all subsequences of a convergent sequence converge to the limit of the sequence, so $p = \alpha = \limsup_\limits{n\to \infty} p_n$.
Similarly $p = \liminf_\limits{n\to \infty} p_n$, hence  $\limsup_\limits{n\to \infty} p_n = \liminf_\limits{n\to \infty} p_n = p$. Therefore by https://proofwiki.org/wiki/Convergence_of_Limsup_and_Liminf, we have $\lim_\limits{n\to \infty} p_n  = p$.
As far as I can tell, none of the lemmas used proof by contradiction.
A: This is based on Ilham's answer. First, we define the real sequence $y_n := d(p_n, p)$ and we have to show $y_n \to 0$.
Long version: (Just to check that no contradiction lurks around)
For every $n \ge 1$, we select $k_n > k_{n-1}$ [with $k_0 = 0$] such that
$$
y_{k_n} \ge \sup\{y_\ell \mid \ell \ge k_{n-1}\} - \frac1n.
$$
Now, there is a subsequence with $y_{k_{n_m}} \to 0$. Thus,
$$
\sup\{ y_\ell \mid \ell \ge k_{n_m-1} \}
\le
y_{k_{n_m}} + \frac1{n_m} \to 0.
$$
Hence, for every $\varepsilon > 0$, there exists $M \ge 1$
with
$$
\sup\{ y_\ell \mid \ell \ge k_{n_m-1} \}
\le
\varepsilon\qquad\forall m \ge M.
$$
Hence,
$$
y_\ell \le \varepsilon \qquad\forall \ell \ge k_{n_M-1},
$$
i.e., $y_\ell \to 0$.
Short version:
Take a subsequence $y_{n_k}$ with $y_{n_k} \to \limsup_{n\to\infty} y_n$. Since a subsequence of the convergent sequence $y_{n_k}$ converges to zero, we have $y_{n_k} \to 0$.
Hence, $\limsup_{n\to\infty} y_n = 0$ and, together with $y_n \ge 0$ for all $n \in \mathbb N$, this gives
$y_n \to 0$.
Conclusio: For me it was interesting to see that this direct argument is also rather short (compared to the usual proof via contradiction).
