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.