Proving $S^3$ and $\mathbb{R}^3$ are not homeomorphic

Question

Prove $S^3$ and $\mathbb{R}^3$ are not homeomorphic.

I've encountered this question on a PhD exam in topology. This is at a level where we are expected to understand cohomology already, so there are already a lot of obvious one line proofs I could give (e.g. they don't have the same homology groups). But this appears among the "give a detailed answer" questions, as opposed to the more computational questions in the latter half. (The heading says to show all work and support all statements to the best of my ability.)

So, I am confused about the level of detail I would have to include here. Is there an obvious choice for how to prove this directly without relying on any one liners that assume higher level stuff?

I realize this question is a little opinion based, but maybe the answer will be unambiguous to those with more experience in topology. How would you answer this question?

Answer 1

$S^3$ is compact whereas $\Bbb R^3$ is not. Thus they cannot be isomorphic.

Answer 2

$S^3$ is compact, while $\mathbb{R}^3$ is not. Since any continuous function $f:S^3\rightarrow \mathbb{R}^3$ maps compact subsets of $S^3$ to compact subsets of $\mathbb{R}^3$, it can't be surjective (or else $f(S^3)=\mathbb{R}^3$ is also compact).