Cauchy sequence is convergent iff it has a convergent subsequence

Certainly not the most elementary proof, but this one feels quite satisfying conceptually: let $(X,d)$ be a metric space and contemplate a Cauchy sequence $\{x_n\}$ with a convergent subsequence, say convergent to $L \in X$. Now consider the completion $\overline{X}$ of $X$: by definition every Cauchy sequence in $\overline{X}$ converges, so our sequence $\{x_n\}$ converges in $\overline{X}$, say to $M$. But then every subsequence also converges to $M$ and thus $M = L$. It follows that the original Cauchy sequence is convergent to $L$!

Added: Really though, the direct proof is also quite simple conceptually: you want to show that the sequence converges to $L$. For any fixed $\epsilon$ we know (i) all pairs of terms with sufficiently large index are within $\frac{\epsilon}{2}$ of each other, and (ii) there is at least one term of the sequence with sufficiently large index which is within $\frac{\epsilon}{2}$ of $L$. Apply the triangle inequality!

Assume that there exists a subsequence $(x_{n_k})$ of $(x_n)$ such that $x_{n_k}\to x$ holds in $X$.Let $\varepsilon>0$ and choose $N$ such that $d(x_{n_k},x)<\varepsilon$ and $d(x_n,x_m)<\varepsilon$ for $n,m>N$. Now, if $n>N$, then $n_k>n>N$,and so $$d(x_n,x)\leq d(x_n,x_{n_k})+d(x_{n_k},x)<2\varepsilon,$$

i.e., $x_n\to x$ in $X$.

Let $x_{n_k}$ $\to$ $L$ .

We need to show that $ \forall$$\varepsilon> 0 $ $\exists N $ such that $d(x_i,L)$ $ $< $\varepsilon$ for $i\geq N$ .

To choose $N$ -- use the Cauchyness of $x_n$ -- $\exists M$ such that $j \geq M$ implies $d(x_M,x_j) < \varepsilon/2$ .

Hence we also have $d(x_M,L)\leq \varepsilon/2$ .

Now take $ N=M$ .

$d(x_i,L) \leq d(x_N,L)+d(x_i,x_N) < \varepsilon/2 + \varepsilon/2 < \varepsilon$ $ \hspace{11mm} \blacksquare$