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The invariance of dimension in plane says that if the set $U \subset \mathbb{R}^n$, when $n \geq 3$, is open and non-empty, then there is no continuous injection $f:U \rightarrow \mathbb{R}^2$.

Surely that can be proven from the general invariance of domain, but is the following planar case enough?

If the set $U \subset \mathbb{R}^2$ is open and non-empty and $f:U \rightarrow \mathbb{R}^2$ is a continuous injection, then the image $f(U)$ is open.

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It's enough. An open $U\subseteq\Bbb R^n$ contains an $S^2$ which is the union of two subsets homeomorphic to the open unit disc. If $f$ is continuous on and injective on the $S^2$ mapping to $\Bbb R^2$, each of these discs maps to an open set. So the image of $S^2$ is open, but also compact...

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