I think I managed to prove:
"If $f:\mathbb{R} \rightarrow \mathbb{R} $ is continuous and open,( i.e., any open $\mathcal{A}\subset\mathbb{R}$ then $f(\mathcal{A})$ is open,) then $f$ is injective."
but I have some doubts especially in the end of the proof. I started with:
Suppose that exists $x\neq y$ such that $f(x)=f(y)$. The restriction of $f$ to $[x,y]$ can not be constant otherwise it would contradict the hypothesis of being an open function. Then exists $z\in(x,y)$ such that $f(z)\neq f(x)$. Suppose that $f(z)>f(x)$ and then prove that $\sup_{w\in (x,y)}\text{$f$}\in f((x,y))$, which is sufficient to contradict the fact of $f$ is an open function.
Consider two sequences $a_{n}$ and $b_{n}$ such that:
$x<a_{n+1}<a_{n}<z<b_{n}<b_{n+1}<y$ and that $a_{n}\rightarrow x$ and $b_{n}\rightarrow y$.
Define $\mathcal{M} _{n}=\max_{w\in [a_{n},b_{n}]}\text{$f$}$, which exists because $f$ is continuous and $[a_{n},b_{n}]$ is compact.
Take $\epsilon$=$\frac{f(z)-f(x)}{2}$ then exists $\delta_{i}>0$, for i=1,2, such that:
- $|w-x|<\delta_{1}$ $\Rightarrow$ $|f(w)-f(x)|<\epsilon$
- $|w-y|<\delta_{2}$ $\Rightarrow$ $|f(w)-f(x)|<\epsilon$
Take $\delta$=min{$\delta_{1},\delta_{2}$}. As $a_{n}\rightarrow x$ and $b_{n}\rightarrow y$ there exists $p_{i}\in\mathbb{N}$ for i=1,2 such that
- $n\geqslant p_{1}$ then $|a_{n}-x|<\delta$
- $n\geqslant p_{2}$ then $|b_{n}-y|<\delta$
Again take $p$=max{$p_{1},p_{2}$} and as $(x,y)=(x,a_{p})\cup[a_{p},b_{p}]\cup(b_{p},y)$ we can conclude:
- $\forall w\in(x,a_{p})\subset(x,x+\delta)$ $\Rightarrow$ $f(w)<f(x)+\epsilon<f(z)$
- $\forall w\in(b_{p},y)\subset(y-\delta,y)$ $\Rightarrow$ $f(w)<f(x)+\epsilon<f(z)$
So exists maximum of $f$ in $(x,y)$ which is attained in $[a_{p},b_{p}]$ and is $\mathcal{M}_{p}$.
Any suggestions would be deeply appreciated.
Here is a proof of the claim that uses the Extreme Value Theorem. I will not spell out complete details because it might be worthwhile to compare the strategy to your proof attempt, especially regarding how much of your proof overlaps with the proof of EVT.
We assume for the sake of a contradiction that there are distinct $a,b$ such that $f(a)=f(b)$. Let $I$ be the open interval $(a,b)$. So $f(I)$ is open by assumption. By the Extreme Value Theorem, $f$ achieves a maximum and minimum value on the closed interval $[a,b]$. Since $f(a)=f(b)$, it follows that at least one of the maximum value or the minimum value is attained in $I$. If the maximum value is attained in $I$, then $f(I)$ contains its supremum, hence is not open. If the minimum value is attained in $I$, then $f(I)$ contains its infimum, hence is not open. So in every case we get a contradiction.