general-topology
Let $U$ be an open sub-set of $\mathbb{R}^2$, and let $f$ be a continuous function $f:U\rightarrow \mathbb{R}^2$.
I'd like to show that $f$ is an open map, given that for each $u\in U$ exists an open set $V_u$ in $U$ ($V_u\subseteq U$) such that $f\uparrow_{V_u}$ is one-to-one.
I'm not quite sure on where to start here?
And assuming I'll prove this, does this mean that $f$ is homeomorphism? Because if $f$ is open so $f^{-1}$ is continuous, right?
A constant function is always continuous since the empty set and the whole space are its only candidates for preimages, and both are open.
A constant function is not necessarily open. It is if and only if in this context $\{y_0\}$ is an open set.
Homeomorphisms are bijective. A constant function can only be bijective if its domain and codomain are both singletons. In that case the function, and also its inverse are automatically continuous, so then indeed we deal with a homeomorphism. So indeed you don't need the topologies here. As said domain and codomain are singletons and on a one-point set there is only one topology.
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 $ V $ be an open subset of $U$. Write $ V $ as a union of open sets $\{V_i\}_{i\in I}$ on which $f$ is 1:1. Apply the Invariance of Domain theorem to each open set $V_i$. Then recall that the union of open sets is open.
Also note that a map like $f$ isn't a homeomorphism, using Chocosup's example or similar. My favorite similar example is to take the open unit square, stretch it to the rectangle $(0,9)\times(0,1)$, and finally wrap it onto (covering $(0,1)\times(0,1)$ twice) $\{(0,3)\times(0,3)\} - \{[1,2]\times[1,2]\}$.
Edit on decomposing $ V $: For each $ v\in V $ let $ \hat V_v $ be a neighborhood with $ f $ 1:1. Then let $ V_v := \hat V_v \cap V $. Because $\mathbb{R}^2$ is reasonable you could slim down the collection $V_v$ to a countable collection, but it isn't needed here.