Suppose the sequence $\{x_n\}$ is Cauchy in $\mathbb{R}$. Prove $\{x_n\}$ is convergent iff it has a convergent subsequence. I have two different approaches in mind. I cannot decide if they are both correct or any of them fails (hopefully not both!).
Method 1:
$\Longrightarrow$ Every Cauchy sequence is bbd. Every bdd sequence has a convergent subsequence.
$\Longleftarrow $ If a subsequence of a Cauchy sequence converges to $x$, then the sequence itself converges to $x$. (A direct proof for this part.)
Method 2:
$\Longrightarrow$ If a sequence converges, then all of its subsequences converge.
$\Longleftarrow $ If a subsequence of a Cauchy sequence converges to $x$, then the sequence itself converges to $x$. (A direct proof for this part.)
Because no one has added an answer yet:
Convergent $\implies$ convergent subsequence is trivial. Every subsequence of a convergent sequence (including the sequence itself) converges to the same limit.
Convergent subsequence + Cauchy $\implies$ convergent is what needs to be shown. Fix $\varepsilon > 0$. Since $\{x_{n}\}$ is Cauchy, I can pick $N_{1} \in \mathbb{N}$ such that $n, m \geq N_{1} \implies |x_{n} - x_{m}| < \varepsilon/2$, and since a subsequence $x_{n_{k}} \to x^{*}$, I can pick $N_{2} \in \mathbb{N}$ such that $k \geq N_{2} \implies |x_{n_{k}} - x^{*}| < \varepsilon/2$. Now, pick $N \geq \max(N_{1}, N_{2})$ (i.e. the larger of the two, so you're big enough to satisfy both conditions). Then for any $n,k \geq N$, we have $$ |x_{n} - x^{*}| \leq |x_{n} - x_{n_{k}}| + |x_{n_{k}} - x^{*}| < \varepsilon/2 + \varepsilon/2 = \varepsilon $$ In essence, we've shown that given $\varepsilon > 0$, we can pick a large enough $N$ such that all terms of the subsequence $\{x_{n_{k}}\}$ are close to the limit $x^{*}$, AND all terms of my sequence are close to each other. This makes sure that all terms of the sequence are close to the limit $x^{*}$.