Prove that if $\sigma : U \subset \mathbb{R}^2\to \sigma(U)\subset \mathbb{R}^3$ is a injective immersion and open then $\sigma$ is a diffeomorphism.
I think that is sufficient prove that $\sigma$ is a locally diffeomorphism.
Let $p \in U$ then $d\sigma$ is injective therefore $T_p U \cong \mathbb{R}^2$. So, $Rank\; d\sigma_p =2$ for every $p \in U$; this implique that $\sigma$ is a parametrization surfaces and exists $\tilde{U}\subset U$ open set with $p \in \tilde{U}$ such that $\sigma: \tilde{U} \to \sigma(\tilde{U})$ is diffeomorphism, that is, $\sigma$ is locally diffeomorphism.
the problem is: I did not use the hypothesis that $\sigma$ is open.
Your argument uses the Inverse Function Theorem, right? But to use that theorem you need two manifolds or, as in this case, two regular parametric surfaces.
Therefore, you should also prove that $\sigma(U)$ is a regular surface. Well, one way to do that is to observe that if $\sigma$ is open, it is also an homeomorphism onto its image, hence a smooth parameterization of it. That's the missing piece.
Besides the formal details, it is always a good idea to visualize the issue. Consider, for an illustration, the Folium of Descartes, with parameterization $$ \Big(\frac{3t}{1+t^3},\frac{3t^2}{1+t^3}\Big) $$
Now, extend it along the $z$ axis as a cylinder (which I will call $F$). We get $$ \sigma(t,z) = \Big(\frac{3t}{1+t^3},\frac{3t^2}{1+t^3},z\Big) $$ This function is an injective immersion. However, its image is not a smooth surface. Coincidentally, $\sigma$ is not open because no open nbh of $(0,0)$ is mapped onto an open nbh of $(0,0,0)\cap F$.