The point here is what means elementary. Many think that a bit of singular homology is not that bad (see Greenberg-Harper's Algebraic Topology textbook, for instance). One can use Mapping Degree Theory as in many books devoted to the topic and use Borsuk-Ulam or else. But I would recommend a carefull reading of
http://terrytao.wordpress.com/2011/06/13/brouwers-fixed-point-and-invariance-of-domain-theorems-and-hilberts-fifth-problem/
There one finds a very nice proof. The heavy part is Brouwer Fixed Point Theorem. Apart from that, some manipulations involving Weierstrass Approximation Theorem and the so-called Little Sard Theorem (in this particular case: a polynomial image of $\mathbb{S}^{n-1}$ in $\mathbb{R}^n$ has empty interior, this disguised in measure zero terms).
First note that the proof of your "lemma" is easy.
For a bijective continuous map $f:X\to Y$ to be a homeomorphism, it is sufficient for $f$ to be a closed/open map, because then
$$
(f^{-1})^{-1}(A) = f(A)
$$
is closed/open for each $A \subset Y$, so that $f^{-1}$ is continuous, whence $f$ is a homeomorphism.
Now note that if $A \subset K$ is closed, where $K$ is compact, then $A$ is compact. Hence, so is $f(A)$. In a Hausdorff space, compact sets are closed, so $f(A)$ is closed, so that $f$ is a closed map.
But this does not proof invariance of domain. To see this, first note that your "proof" would note use the fact that $U \subset \Bbb{R}^n$ and $f : U \to \Bbb{R}^n$ (note that the dimensions match). But without matching dimensions, the theorem is not valid, as the following counterexample (taken from http://en.wikipedia.org/wiki/Invariance_of_domain#Notes) shows:
$$
f : (-1.1\, , \, 1) \to \Bbb{R}^2, x \mapsto (x^2 - 1, x^3 - x).
$$
The image of this function (also taken from the same post) is
It is an easy exercise to show that $f$ is not a homeomorphism onto its image although it is continuous and injective.
The problem here is that the claim you get is only that each restricted map $f|_K : K \to f(K)$ is a homeomorphism for $K \subset U$ compact. But this only gives you continuity of $f^{-1}|_{f(K)}$. But this does not entail continuity of $f^{-1}$ (as the example shows).
Best Answer
Let's call a topological space $X$ domain invariant, if it fullfills the usual condition of the theorem of domain invariance, i.e. if $U, V$ are homeomorphis subsets of $X$, and $U$ is open, then also $V$ is open.
$X$ fullfills (*), if every continuous bijection $f: X \rightarrow X$ is a homeomorphism (sorry, no better notation came up to my mind).
By a standard argument we have: If X is doamin invariant, T2 and locally compact, then $X$ fullfills (*).
Proof. Let $f: X \rightarrow X$ be a continuous bijection. Of course, it suffices to show that $f$ is open: Let $U$ be open in $X$, $x \in U$. By local compactness pick an open $V$ such that $x \in V \subset \overline{V} \subset U$ and $\overline{V}$ compact. Then $f|\overline{V}: \overline{V} \rightarrow f(\overline{V})$ is continuous, bijective, hence a homeomorphism. Thus, $f|V: V \rightarrow f(V)$ is a homeomorhism, and $ V \cong f(V)$. Since $X$ is domain invariant, $f(V)$ is open. Therefore $f(U)$ is open.
This might be interpreted as "$\Rightarrow$" of the above statement (if $X$ is T2, locally compact). However, "$\Leftarrow$" is not true in general: Of course, if $X$ is compact, T2 then $X$ fullfills (*), but need not be domain invariant, as the unit interval shows.