Well, I might recast your proofs of the first two cases as follows:

Suppose that $\mathbb{R}^n$ and $\mathbb{R}^m$ are homeomorphic. Then for any $P \in \mathbb{R}^n$, there must exist a point $Q \in \mathbb{R}^m$ such that $\mathbb{R}^n \setminus \{P\}$ and $\mathbb{R}^m \setminus \{Q\}$ are homeomorphic. (Since the homeomorphism group of $\mathbb{R}^m$ acts transitively, really any point $Q$ is okay, but I'm trying to be both simple and rigorous.)

Now your proof when $n = 1$ is equivalent to the observation that $\pi_0(\mathbb{R}^1 \setminus \{P\})$ is nontrivial, while $\pi_0(\mathbb{R}^n \setminus \{Q\})$ is zero for all $n > 1$. (Note that $\pi_0(X)$ is in bijection with the set of path-components of $X$, so really we are using that $\mathbb{R}^1$ minus a point is path-connected and $\mathbb{R}^n$ minus a point is not, for $n > 1$.)

When $n =2$, your proof is literally that $\pi_1(\mathbb{R}^2 \setminus \{P\})$ is nonzero whereas $\pi_1(\mathbb{R}^n \setminus \{Q\})$ is zero for all $n > 2$. It is a little questionable to me whether the fundamental group counts as "elementary" -- you certainly have to learn about homotopies and prove some basic results in order to get this.

If you are okay with such things, then it seems to me like you might as well also admit the higher homotopy groups: the point here is that

for all $m \in \mathbb{Z}^+$, $\pi_{m-1}(\mathbb{R}^m \setminus \{P\}) \cong \mathbb{Z}$ whereas for $n > m$, $\pi_{m-1}(\mathbb{R}^n \setminus \{Q\}) = 0$.

If I am remembering correctly, the higher homotopy groups are introduced very early on in J.P. May's *A Concise Course in Algebraic Topology* and applied to essentially this problem, among others.

[By the way, if we are okay with homotopy, then we probably want to replace $\mathbb{R}^n \setminus \{P\}$ with its homotopy equivalent subspace $S^{n-1}$ throughout. For some reason I decided to avoid mentioning this in the arguments above. If I had, it would have saved me a fair amount of typing...]

**Added**: Of course one could also use homology groups instead, as others have suggested in the comments. One might argue that homotopy groups are easier to *define* whereas homology groups are easier to *compute*. But this one computation of the "lower" homotopy groups of spheres is not very hard, and my guess is that if you want to start from scratch and prove everything, then for this problem homotopy groups will give the shorter approach.

As to why the problem is hard to solve in an elementary way: the point is that the two spaces $\mathbb{R}^m \setminus \{P\}$ and $\mathbb{R}^n \setminus \{Q\}$ look the same when viewed from the lens of general topology. [An exception: one can develop **topological dimension theory** to tell them apart. For this I recommend the classic text of Hurewicz and Wallman. Whether that's "more elementary", I couldn't say.] In order to distinguish them for $m,n$ not too small, it seems that you need to develop various notions of the "higher connectivities" of a space, which leads inevitably to homotopy groups and co/homology groups.

Another alternative is to throw out homeomorphism and look instead at diffeomorphism. This puts a reasonable array of tools from differentiable topology and manifold theory at your disposal rather quickly (see e.g. Milnor's book *Topology from a differentiable viewpoint*). It is not hard to show that the dimension of a manifold is a diffeomorphism invariant! So maybe the subtlety comes from insisting on working in the topological category, which often turns out to be more difficult than working with topological spaces with nice additional structures.