Let $g:\mathbb R^n\rightarrow \mathbb R$ continuous and injective. Show that $g(U)$ is open if $U\subset \mathbb R^n$ open. 
Let $g:\mathbb R^n\rightarrow \mathbb R$ continue and injective. Show that $g(U)$ is open if $U\subset \mathbb R^n$ open. Let $U\subset \mathbb R^n$ open and $g(a)\in g(U)$. 

I naturally have to show that $$]g(a)-\varepsilon, g(a)+\varepsilon[\subset g(U)$$ for a certain $\varepsilon>0$.
We have that $a\in g^{-1}(g(U))=U$ by injectivity. Since $U$ is open, there is a $\delta>0$ s.t.$$]a-\delta,a+\delta[\subset U.$$
Let $x\in]a-\delta,a+\delta[$ and let $\varepsilon>0$. By continuity, there is $0<\eta <\delta$ such that for all $x\in B_\eta(a)$ $$g(x)\in ]g(a)-\varepsilon,g(a)+\varepsilon[.$$
From here, I don't see how to conclude. 
 A: If $n=1,$ this is well known. If $n>1,$ there are no such maps. (Why? Let $a\in U$ and consider two different lines through the point $a.$)
A: @zhw already adressed your question. I'll provide an answer under a different POV and which also contains the case $n=1$.
We assume $n>1$ (Read OBS at the end).  
Let $g: \mathbb{R}^n \rightarrow \mathbb{R}$ be continuous and injective, and denote by $K_i=[-i,i]^n$. Since $K_1$ is compact and connected, its image is a set of the form $[a_1,b_1]$. Since $g$ is injective and $K_{i+1} \setminus K_{i}$ is connected, $g(K_{2})$ must preserve some endpoint,  being either $[a_1,b_2]$ or $[a_2,b_1]$ for some $b_2>b_1$ or $a_2 < a_1$. Suppose wlog that $g(K_2)=[a_1,b_2]$. Therefore, $g({K_2 \setminus K_1})=[b_1,b_2]$. By same reasoning as before, $g(K_3)$ must be of the form $[\eta, b_2]$ or $[a_1,\eta']$. However, if it was of the first form, then the image $g(K_3 \backslash K_1)$ would be $[\eta, a_1] \cup [b_1,b_2]$. With a simple inductive argument, we conclude that $g(K_i)=[a,b_i]$, where $b_n$ is an increasing sequence.
It is now easy to show that $g(\mathbb{R}^n)=[a,\beta)$, where $\beta$ is a real number or infinity. Either way, by what was shown above, we can define a continuous extension of the function $g$, which we will call $\tilde{g}$ from the-point compactification $C$ of $\mathbb{R}^n$ to the one-point compactification $C'$ of $[0,\beta)$ by $\tilde{g}=g$ in $\mathbb{R}^n$ and $\tilde{g}(\infty)=\infty$. $\tilde{g}$ is obviously an injective map. But since $\tilde{g}$ is a continuous bijective map from a compact set, it is a homeomorphism. 
Therefore, since $C \cong S^n$ and $C' \cong [0,1]$ , we have that $S^n \cong [0,1]$, which is a contradiction. Therefore, there does not exist a map satisfying your conditions.
OBS: Note that in our argument on the first paragraph we used the fact that $K_i \backslash K_{i-1}$ is connected, a fact which is only valid if $n>1$. The preceding arguments can be adapted to $n=1$ in order to show the case $n=1$. The big change is the fact that $g(K_i)$ can, and will, be of the form $[a_i,b_i]$ with $a_i$ decreasing and $b_i$ increasing. Thus, it will easily follow that the image is open (equal to $(a,b)$, where $a =\lim a_i$, $b=\lim b_i$). You can even show it is a homeomorphism with its image, but you'll then have to perform the two-point compactification for that, with slight changes on strategy.
A: A continue map transforms a connected set to a connected set, if U is open x in U, if you consider a connected segment s which is contained in U and contains x in its interior, g(s) is connected, thus is an interval of R which is not reduced to a point since g is injective, thus g(U) is open. Since g(x) is in the interior of g(s) since g is injective.
