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).
Since $f:U\to V$ is a homeomorphism, $\space f$ must have a continuous inverse $\space f^{-1}: V \to U$. Now we use the fact that $f^{-1}$ is continuous, which means any open subset $W \subset U$ is such that $(f^{-1})^{-1}(W) = f(W) \subset V$ is open, completing the proof.
The notation isn't great, but basically use the "inverse image of an open set is open" definition of continuity to see that the inverse of $f^{-1}$ (which is $f$) produces open images in $V$
Best Answer
It sufficies to show that for any $a,b\in \mathbb R$ with $a<b$, $f(a,b)$ is an open interval.
Since $f$ is continuous, $f[a,b]$ is compact and connected in $\mathbb R$, hence $f[a,b]=[c,d]$, for some $c,d\in \mathbb R$, let us see $f(x)\neq d$ for all $x\in (a,b)$. Suppose $d=f(x)$ for some $x\in (a,b)$, then since $f$ is injective $f(a)<d$ and $f(b)<d$, hence we can choose $r\in (c,d)$ such that $f(a)<r$ and $f(b)<r$. Then by the intermediate value theorem there is $y\in (a,x)$ such that $f(y)=r$ and there exists $z\in(x,b)$ with $f(z)=r$, contradicting the injectivity of $f$. Thus $f(x)\neq d$ for all $x\in (a,b)$. Similarly $f(x)\neq c$ for all $x\in (a,b)$. Hence $f(a,b)=(c,d)$, since $f(a,b)$ is connected and $f[\{a,b\}]=\{c,d\}$.