f is open map and continuous, let f:R->R
Let the inverse of f is F.so F is continuous from f(R) to R.
Now lets assume there exists distinct x,y in R such that f(x)=f(y).
WLG lets assume f(x)=f(y)=c for some c in R.
F(x) is not continuous at c as x,y are distinct and hence the contradiction.
Suppose $f(x_1) = f(x_2)$. If $x_1<x_2$, then $f(x_1) < f(x_2)$, and similarly, if $x_1>x_2$, then $f(x_1) > f(x_2)$. It follows that $x_1 = x_2$ and hence $f$ is injective.
Since $f$ is injective, it is invertible on its range $B=f(\mathbb{R})$. Let $f^{-1}$ denote this inverse. Suppose $x_1 < x_2$, and suppose $y_1 = f^{-1}(x_1)$, $y_2 = f^{-1}(x_2)$. We must have $y_1 \neq y_2$, otherwise we would have $f(y_1) = f(y_2)$, a contradiction. We must have $y_1 < y_2$, because otherwise, if $y_1>y_2$, then $f(y_1) > f(y_2)$, again a contradiction. Hence $f^{-1}$ is strictly increasing.
Let $x \in B$, and $\epsilon>0$. Let $y = f^{-1}(x)$. Let $x_+ = f(y+\epsilon)$, $x_- = f(y-\epsilon)$. Note that if $x_- < z < x_+$, then since $f^{-1}$ is strictly increasing, we have $y-\epsilon < f^{-1}(z) < y+\epsilon$.
To finish, let $\delta = \min(x_+-x, x-x_-)$. Then if $|z-x| < \delta$, we have $|f^{-1}(z)-f^{-1}(x)| < \epsilon$, hence $f^{-1}$ is continuous.
Aside: I left out a minor detail above. Since $f$ is strictly increasing, it follows that $B$ is open (since $\mathbb{R}$ is open). When showing continuity of $f^{-1}$, the $\epsilon$ must be taken small enough so that $B(x,\epsilon) \subset B$.
Best Answer
Let $M:=\{(x,y)|x<y\}$. Define $g\colon M\to \mathbb R$ by $g(x,y):=f(x)-f(y)$. Clearly $g$ is continuos and $M$ is connected. Hence $g(M)$ is connected, i.e., an interval of $\mathbb R$ which does not contain $0$ as $f$ is injective. That is, $g$ is either strictly negative or strictly positive, hence $f$ is strictly increasing resp. decreasing.