At first I spent a lot of time looking for counterexamples because I had never seen such a claim that M.P. implies O.M.T.. But later I realized the claim might be true, so I just had a try and proved it! I checked my steps and I think the proof is correct. But I still feel unsafe, so I asked this question.
There is a same question in math stackexchange. Here is the link: Prove the open mapping theorem by using maximum modulus principle. I found the answer unsatisfactory, at least it didn't provide a counterexample. I posted my proof right behind the original answer.
The following is the proof. Just for convenience for the reader, I copied it again here.
Suppose $f$ is the non constant continuous function that satisfies the maximal principle property. If open mapping theorem is not true, then $f$ maps an interior point $x$ of a small closed neighborhood $D$ to the point $f(x)$ which is on the boundary of $f(D)$.
Then using translation $g$ to compose with $f$, to make sure $|g(f(x)|$ has the maximal moduli over $D$. We can do that because $f(D)$ is compact, so there exits $z_0$ in $f(D)$ such that dist($f(x), z_0$) is equal to the maximum of the distance from $f(x)$ to all points in $f(D)$, then consider $g(z)=z-z_0$. Now $g(f(x))$ achieves its local maximal point over $D$ at $x$, by hypothesis, $g(f(x))$ is constant function on $D$ since maximal principle is preserved under translation, i.e., a non constant function still gets its local maximal modulus on the boundary of the local neighborhood after translation.
Therefore, $f$ is a constant function on $D$. Using typical method (for example, Lebesgue lemma) for path-connectness of a domain, we say that $f$ is constant on its domain.
So, it seems that maximal modulus principle implies open mapping theorem for any continuous functions, but I've never seen such a claim before and nobody mentioned it. Any comments would be appreciated.