Suppose $\mathbb{Q_p} $ is the fraction field of $\mathbb{Z_p}$ ($p$-adic integers) i.e.
$$\mathbb{Q_p} = \left\lbrace\frac{x}{y} \space \bigg{|} \space x,y \in \mathbb{Z_p} , y\neq 0 \right\rbrace$$
Now with respect to the topology defined by $d(x,y) = e^{-v_p(x-y)}$ ($v_p$ is the $p$-adic valuation) , we need to show that $\mathbb{Q_p} $ is locally compact.
Any suggestions?
Every point has a fundamental system of neighborhoods given by $\{x+p^n\Bbb Z_p\}_{n\in\Bbb N}$ which are compact.
In essence this is just the fact that $p^n\Bbb Z_p$ is a fundamental system of compact neighborhoods of $0$: since we are in a vector space--really a topological group is enough, but not everyone is familiar with structures of that generality--we can translate these sets anywhere to form a compact (and open) neighborhood of any point.
If you have any trouble seeing this recall that open balls generate the topology and all open balls are of the form $x+p^n\Bbb Z_p$ for some $n\in\Bbb Z$.
The accepted answer is not complete because it does not argue why $p^n\mathbb{Z}_p$ is compact. I think the following idea deal with this:
To prove that $\mathbb{Q}_p$ is locally compact is enough to prove that $\mathbb{Z}_p$ is compact. For this, thinking in the $p$-adic integers as sequences we have $$\mathbb{Z}_p=\{(a_0,a_1,a_2,...)\mid a_n\in \mathbb{Z}/p\mathbb{Z} \}=(\mathbb{Z}/p\mathbb{Z})^\mathbb{N}$$
Now, by Tychonoff $(\mathbb{Z}/p\mathbb{Z})^\mathbb{N}$ is compact with the product topology and also is easy to see that $x+p^n\mathbb{Z}_p$ is an open set of this topology. So every cover of $\mathbb{Z}_p$ with sets of the form $x+p^n\mathbb{Z}_p$ will have a finite subcover and then $\mathbb{Z}_p$ is compact.