In a metric space, is every convergent sequence bounded? In $\mathbb{R}$ and $\mathbb{R}^p$, this is true, but is it true in every metric space?
I suppose not, but what other condition would I have to put on the metric space in order for it to have this property?
For clarity, a bounded subset $W$ of a metric space $V,d$ is a subset of $W$ with this property:
$$\exists M\in\mathbb{R}_0^+,\forall v,w \in V:\ d(v,w) < M$$

Yes, it’s true in every metric space. In fact, a stronger statement is true: every Cauchy sequence is bounded. Let $\langle x_n:n\in\Bbb N\rangle$ be a Cauchy sequence in a metric space $\langle X,d\rangle$. Since the sequence is Cauchy, there is an $m\in\Bbb N$ such that $d(x_k,x_\ell)<1$ whenever $k,\ell\ge m$. In particular, $d(x_m,x_n)<1$ for all $n\ge m$. Now let $\alpha=\max\{d(x_m,x_n):n<m\}$; then $d(x_n,x_m)<\alpha+1$ for each $n\in\Bbb N$, so the sequence is bounded.
Added: Specifically, for all $k,\ell\in\Bbb N$ we have
$$d(x_k,x_\ell)\le d(x_k,x_m)+d(x_m,x_\ell)<2\alpha+2\;.$$
