A elementary proof can be found in the following article:
An interesting proof on the nonexistence of a continuous function between $\mathbb{R}^2$ and $\mathbb{R}^n$ for $n \neq 2$, by F. Malek, H. Daneshpajouh, H.R. Daneshpajouh and J. Hahn.
I'm just expanding on Pink Panther's comment. If I understood correctly, your idea was to
- consider a space $Y$ such that $Y \equiv \mathbb{R}$ and its compactification is $Y^* \equiv \mathbb{S}^1$, and then
- use that $X \equiv Y$ implies $X^* \equiv Y^*$.
Moreover, since it is clear that the compactification of $\mathbb{S}^1 \setminus \{(0,-1)\}$ is the $1$-sphere, it suffices to provide an isomorphism from this space to $\mathbb{R}$.
So, what you can do to generalize this is to find $Y_n$ such that $Y_n \equiv \mathbb{R}^n$ and $Y_n^* \equiv \mathbb{S}^n$. By the argument above, it suffices to find a homeomorphism
$$
p : \mathbb{S}^n \setminus \{N\} \to \mathbb{R}^n
$$
for some point $N$ of the $n$-sphere. Here's where the stereographic projection comes into play.
Geometrically, imagine the $n$-sphere embedded into $\mathbb{R}^{n+1}$ and for convinience, pick $N = e_{n+1}$ the north pole. Then, for any other point $q$ of the sphere, the line $\vec{Nq}$ intersects the plane $\Pi := \{x_{n+1} = 0\}$ of the 'floor' exactly once. Call this point $p(q)$. Moreover, one can see that this mecanism reaches every point of $\Pi$, which can be identified with $\mathbb{R}^n$.
Concretely now, via a calculation that I ommit (but I encourage you to figure out yourself before checking the literature), we can define
$$
\begin{align}
p :\ & \mathbb{S}^n \setminus \{N\} \to \mathbb{R}^n\\
&(x,t) \mapsto \frac{1}{1-t}x
\end{align}
$$
Note that this is well defined, since the only point in the sphere with $t = 1$ is the north pole $N = (0,\dots, 0,1)$ which we have excluded.
You can check that this is a homeomorphism with inverse
$$
\begin{align}
p :\ & \mathbb{R}^n \setminus \{N\} \to \mathbb{S}^n\\
&y \mapsto \frac{1}{\|y\|^2+1}(2y,\|y\|^2-1)
\end{align}
$$
Best Answer
There are reasonably accessible proofs that are purely general topology. First one needs to show Brouwer's fixed point theorem (which has an elementary proof, using barycentric subdivion and Sperner's lemma), or some result of similar hardness. Then one defines a topological dimension function (there are 3 that all coincide for separable metric spaces, dim (covering dimension), ind (small inductive dimension), Ind (large inductive dimension)), say we use dim, and then we show (using Brouwer) that $\dim(\mathbb{R}^n) = n$ for all $n$. As homeomorphic spaces have the same dimension (which is quite clear from the definition), this gives the result. This is in essence the approach Brouwer himself took, but he used a now obsolete dimension function called Dimensionsgrad in his paper, which does coincide with dim etc. for locally compact, locally connected separable metric spaces. Lebesgue proposed the covering dimension, but had a false proof for $\dim(\mathbb{R}^n) = n$, which Brouwer corrected.
One can find such proofs in Engelking (general topology), Nagata (dimension theory), or nicely condensed in van Mill's books on infinite dimensional topology. These proofs do not use homology, homotopy etc., although one could say that the Brouwer proof of his fixed point theorem (via barycentric division etc.) was a precursor to such ideas.