Problem 15 chapter 4 from baby Rudin Call a mapping of $X$ into $Y$ open if $f(V)$ is an open set in $Y$ whenever $V$ is an open set in $X$.
Prove that every continuous open mapping of $\mathbb{R}^1$ into $\mathbb{R}^1$ is monotonic.
Proof: Let $f$ is not monotonic function. Then exists $a<b<c$ such that $f(a)>f(b)$ and $f(c)>f(b)$. Hence $[a,c]$ is compact then continuous $f$ attains it's minimum at q, i.e. $\text{argmin}_{[a,c]}f=q.$ But $f(a)>f(b)\geqslant f(q)$ and $f(a)>f(b)\geqslant f(q)$ thus $\text{argmin} _{(a,c)}f=q.$ 
Let $(q-\delta,q+\delta)$ some neighborhood of $q$ and it's an open set in $\mathbb{R}^1$ and $f$ is an open mapping then $f((q-\delta,q+\delta))$ must be an open, but $f((q-\delta,q+\delta))$ is not open because the last set contains no neighborhood of $q$.
Is my proof true?
 A: [edit] If you edit the proof, my answer becomes a little silly. Hence I have copied your original proof here: this is what I am commenting on.

Proof: Let $f$ is not monotonic function. Then exists $a<b<c$ such that $f(a)>f(b)$ and $f(c)>f(b)$. Hence $[a,c]$ is compact then continuous $f$ attains it's minimum at q, i.e. $\text{min}_{[a,c]}f=q.$ But $f(a)>f(b)\geqslant f(q)$ and $f(a)>f(b)\geqslant f(q)$ thus $\text{min} _{(a,c)}f=q.$
Let $(q-\delta,q+\delta)$ some neighborhood of $q$ and it's an open set in $\mathbb{R}^1$ and $f$ is an open mapping then $f((q-\delta,q+\delta))$ must be an open, but $f((q-\delta,q+\delta))$ is not open because the last set contains no neighborhood of $q$.

It is essentially right. Some points-

*

*It is not true that you will get a minimum, since it could be a maximum. Of course this follows, but you didn't say so.

*When you say $\min_{[a,c]}f = q$, what you actually want to say is $\text{argmin}_{[a,c]} f = q$

*You repeated $f(a) > f(b) \geq f(q)$ twice; I assume one of them should be $f( c ) > f(b) \ge f(q)$.

*You did not show that $f((q-δ,q+δ))$ is not open: what is true is that it does not have a neighbourhood around of $f(q)$, and also only for $δ$ small enough so that $(q-δ,q+δ)\subset [a,c]$.


Lemma.
Say a function f is non-monotone if there exist $a<b<c$ such that either $f(b) > max(f(a),f(c )) $ or $f(b) < min(f(a),f(c ))$. Then  f is non-monotone ⇔ it is not monotone increasing and it is not monotone decreasing.
Proof. "$\implies$" is clear. Suppose f is not non-monotone. Then for all triples $a<b<c$, $f(b)$ is in between the values $f(a)$ and $f(b)$.
Suppose $f(a)\le f(b)\le f(c)$ for one triple. We need to show that for any $x<y, f(x)\le f(y)$.
There are a few cases to consider, but in each case you just need to draw it out and show that if $f(x)>f(y)$ we get a 'V' shape in the graph. I will illustrate one-
$x<y \le a < b < c$. We must have $f(y)$ in between $f(x)$ and $f(b)$. But if $f(x)>f(y)>f(b)$, then $f(b)<min(f(y),f(c ))$, which contradicts the assumption that $f$ is not non-monotone. Thus $f(x)\le f(y) \le f(a)$ as needed.
The other cases are clear by analogy. We can also proceed similarly if we have a decreasing triple instead of an increasing one to show that $f$ is monotone decreasing. QED
