[proofplan]
We argue by contradiction. If $\mathbb{R}^n \cong \mathbb{R}^m$ via a homeomorphism $h$, then removing the origin and its image produces a homeomorphism $\mathbb{R}^n \setminus \{0\} \cong \mathbb{R}^m \setminus \{0\}$. Since $\mathbb{R}^d \setminus \{0\} \simeq S^{d-1}$ for all $d \geq 1$, this yields $S^{n-1} \simeq S^{m-1}$. But [spheres of different dimensions are not homotopy equivalent](/theorems/2243) when $n \neq m$, giving the desired contradiction.
[/proofplan]
[step:Assume $h: \mathbb{R}^n \to \mathbb{R}^m$ is a homeomorphism and restrict to punctured spaces]
Suppose for contradiction that there exists a homeomorphism $h: \mathbb{R}^n \to \mathbb{R}^m$ with $n \neq m$. Let $p := h(0) \in \mathbb{R}^m$. The restriction of $h$ to $\mathbb{R}^n \setminus \{0\}$ gives a homeomorphism
\begin{align*}
h|_{\mathbb{R}^n \setminus \{0\}}: \mathbb{R}^n \setminus \{0\} \xrightarrow{\;\cong\;} \mathbb{R}^m \setminus \{p\}.
\end{align*}
Since translation $x \mapsto x - p$ is a homeomorphism of $\mathbb{R}^m$ sending $p$ to $0$, we obtain $\mathbb{R}^m \setminus \{p\} \cong \mathbb{R}^m \setminus \{0\}$. Composing, we have
\begin{align*}
\mathbb{R}^n \setminus \{0\} \cong \mathbb{R}^m \setminus \{0\}.
\end{align*}
[/step]
[step:Identify $\mathbb{R}^d \setminus \{0\}$ with $S^{d-1}$ up to homotopy equivalence]
For any $d \geq 1$, the inclusion $i: S^{d-1} \hookrightarrow \mathbb{R}^d \setminus \{0\}$ and the radial projection
\begin{align*}
r: \mathbb{R}^d \setminus \{0\} &\to S^{d-1} \\
x &\mapsto \frac{x}{|x|}
\end{align*}
satisfy $r \circ i = \operatorname{id}_{S^{d-1}}$. The composition $i \circ r$ is homotopic to $\operatorname{id}_{\mathbb{R}^d \setminus \{0\}}$ via the straight-line homotopy
\begin{align*}
F: [0, 1] \times (\mathbb{R}^d \setminus \{0\}) &\to \mathbb{R}^d \setminus \{0\} \\
(t, x) &\mapsto (1 - t)x + t \frac{x}{|x|}.
\end{align*}
This is well-defined: for each $x \neq 0$ and $t \in [0, 1]$, the point $F(t, x) = ((1-t) + t/|x|) \cdot x$ is a positive scalar multiple of $x$ (since $(1-t) + t/|x| > 0$), hence nonzero.
Therefore $S^{d-1}$ is a deformation retract of $\mathbb{R}^d \setminus \{0\}$, giving $\mathbb{R}^d \setminus \{0\} \simeq S^{d-1}$.
[/step]
[step:Derive the contradiction from the non-homotopy-equivalence of spheres]
Combining the previous two steps:
\begin{align*}
S^{n-1} \simeq \mathbb{R}^n \setminus \{0\} \cong \mathbb{R}^m \setminus \{0\} \simeq S^{m-1}.
\end{align*}
Since homeomorphisms are in particular homotopy equivalences, and homotopy equivalence is transitive (as an equivalence relation by [Homotopy Equivalence is an Equivalence Relation](/theorems/1874)), this gives $S^{n-1} \simeq S^{m-1}$.
However, $n \neq m$ implies $n - 1 \neq m - 1$, so by [Spheres of Different Dimensions are Not Homotopy Equivalent](/theorems/2243), $S^{n-1} \not\simeq S^{m-1}$. This is a contradiction.
Therefore no such homeomorphism $h$ exists, and $\mathbb{R}^n \not\cong \mathbb{R}^m$.
[guided]
The argument reduces a question about homeomorphisms of Euclidean spaces to a question about homotopy types of spheres, which we can answer using homology.
The key idea is that a homeomorphism $h: \mathbb{R}^n \to \mathbb{R}^m$ restricts to a homeomorphism of the "punctured" spaces $\mathbb{R}^n \setminus \{0\} \cong \mathbb{R}^m \setminus \{p\}$ (where $p = h(0)$). Translation in $\mathbb{R}^m$ takes care of the basepoint issue: $\mathbb{R}^m \setminus \{p\} \cong \mathbb{R}^m \setminus \{0\}$.
Why puncture? Because $\mathbb{R}^n$ and $\mathbb{R}^m$ are both contractible for any $n, m$, so their homology groups are identical and cannot distinguish them. Removing a point introduces nontrivial topology: $\mathbb{R}^d \setminus \{0\}$ deformation retracts onto $S^{d-1}$, and spheres of different dimensions have different homology groups. This is a recurring technique in algebraic topology: to distinguish spaces, modify them in a way that reveals hidden structure.
The deformation retraction $\mathbb{R}^d \setminus \{0\} \simeq S^{d-1}$ works by radial projection: every nonzero vector $x$ can be continuously normalized to $x / |x| \in S^{d-1}$. The straight-line homotopy from $x$ to $x / |x|$ stays in $\mathbb{R}^d \setminus \{0\}$ because the intermediate points are positive multiples of $x$.
The final contradiction comes from [Spheres of Different Dimensions are Not Homotopy Equivalent](/theorems/2243), which in turn relies on the computation $H_{n-1}(S^{n-1}) \cong \mathbb{Z} \neq 0 = H_{n-1}(S^{m-1})$ (for $n \neq m$ with appropriate case analysis).
[/guided]
[/step]
