Consider the image of an open set $(a,b)$ under the open and continuous mapping $f$. We show, $f$ cannot have any extremum in $(a,b)$.
We know, connected sets are mapped to connected sets under a continuous map. Hence, $f[(a,b)]=(c,d)$ is connected (and open). Suppose, at $\xi \in (a,b)$, $f(\xi)=\sup_{(a,b)} f=M$ (or $\inf_{(a,b)}f=m$). [Being a continuous function, $f$ must attain its supremum/infimum.]
Hence, the image of the set $(a,b)$ under $f$ becomes $(c,M]$ or $[m,d)$, which is a contradiction of the fact that $f$ maps open sets to open sets. Therefore, in any open interval, the function cannot attain glb/lub at an interior point. So, we conclude that the inf and sup are at the end points, i.e. $\sup_{[a,b]}f=f(a)$ or $f(b)$.
Hence the theorem.
Is the proof valid? I am aware of the duplicates. I just want this method verified.