I am trying to prove that compact sets are bounded, using the sequential definition of compactness. Is the following correct?
Suppose $K$ is a compact subset of a metric space $(X,d)$. Suppose, for a contradiction, that $K$ is not bounded. Then, for all $r>0$ and $x \in X$, there exists some $x_0 \in K$ s.t. $d(x, x_{0})>r$. Using this fact, construct the following sequence. Choose some arbitrary point in $K$, and make it $a_{1}$. For $r=1$, we can find some $x\in K$ such that $d(a_{1}, x)>1$. Set $a_{2}:=x$. Generally, choose $a_{n}$ such that $d(a_{1}, a_{n})>n$. This completes the construction of $ \{a_{n}\}$ (which, note, is wholly contained in $K$). I now show that any subsequence of $\{a_{n}\}_{n \in \mathbb{N}}$ is not Cauchy (and so not convergent—contradicting the assumption that $K$ is compact).
Consider an arbitrary subsequence of $\{a_{n}\}$, $\{a_{n_{k}}\}$. Set $\epsilon=1$, and consider an arbitrary $P\in \mathbb{N}$. Choose some arbitrary $k_{1}>P$. By the above, we have that $d(x_{1}, a_{k_{1}})>k_{1}$. Now, choose some $k_{2}$ such that $k_{2}\geq 1+ d(a_{k_1},a_{1}).$ Then, $d(a_{k_2}, a_{1})>1+d(a_{k_1},a_1)$. That is, $d(a_{k_2}, a_{1})-d(a_{k_1},a_1)>1$. It follows that $|d(a_{k_2}, a_{1})-d(a_{k_1}, a_1)|>1$. By the reverse triangle inequality, $d(a_{k_2}, a_{k_1})\geq|d(a_{k_2}, a_{1})-d(a_{k_1},a_1)|>1$. Thus, $\{a_{n_k}\}$ is not Cauchy.
The reasoning is correct. The last part (after the construction) can be substantially shorter. Namely, if $\{a_{n_k}\}$ is a convergent (to $a$) subsequence then from the continuity of the distance function we have $$\infty\leftarrow n_k < d(a_1,a_{n_k}) \to d(a_1,a),$$ a contradiction.