Does finite covering dimension imply local compactness?

Question

I have a space which is not locally compact and I'm trying to see if I can say anything about the dimension of the space. I suspect that it is not finite dimensional but I have thus far been unable to prove it. There doesn't seem to be an obvious connection between finite dimension and local compactness, but I feel like I might be missing something.

Answer 1

$\mathbb{Q}$ has a base of clopen sets and is zero-dimensional but is also a standard example of a non-locally compact space.

Similarly, $\mathbb{Q}\times [0,1]^n$ has Lebesgue covering dimension $n$ but is not locally compact.

Answer 2

Uhm, how about the rationals? Not locally compact, but dimension is 1.

Edit: as observed by Jeremy in the comments, the topological dimension of $\mathbb{Q}$ is 0, while 1 is the dimension as a vector space (over $\mathbb{Q}$, of course). In any case, it is finite.