Prove that $f$ continuous if and only if $f([a,b])=[c,d]$ for $f:[a,b]\rightarrow \mathbb{R}$ strictly increasing and $a,b,c,d \in \mathbb{R}$

Question

So I'm trying to prove the following:

Suppose $a,b\in\mathbb{R},\quad a<b$ and $f:[a,b]\rightarrow \mathbb{R}$ a strictly increasing function.

Prove that $f$ continuous if and only if there exist $c,d\in\mathbb{R}$ with $f([a,b])=[c,d]$

My attempt: Suppose $f([a,b])=[c,d]$ with $c,d$ like above. Then $f$ is invertible because it is both injective (strictly increasing) and surjective. Consider a point $f(p)\in [c,d]$, and let $\epsilon>0$. Pick $f(x)\in [c,d]\setminus f(p) $ such that $d(f(x),f(p))<\epsilon$ and assume without loss of generality that $f(x)<f(p)$. (The other case is entirely analogous.) Then there exists some $f(x)<f(x_0)<f(p)$ and because $f(x_0)\in [c,d],\quad x_0\in f^{-1}([c,d])=[a,b]$. Here we'll use the strictly increasing nature of $f$ to conclude $x<x_0<p$ with $x,x_0,p\in[a,b]$ Thus, if we let $$d(x_0,p)<d(x,p),\qquad d(f(x),f(p))<\epsilon$$ Which is exactly the definition of continuity.

Conversely, suppose $f$ is continuous on $[a,b]$. Knowing that the image of a closed set is against closed under a continuous function, $f([a,b])$ is closed. Suppose $t\in [a,b]$, then because $f$ is strictly increasing, $f(a)\leq f(t)\leq f(b)$, and $f([a,b])\subset [f(a),f(b)]$. Consider $f(s)\in[f(a),f(b)]$, then $f(a)\leq f(s) \leq f(b)$ so $a\leq s \leq b$ and $s\in [a,b]$. Therefore $[f(a),f(b)]\subset f([a,b])$ and $f([a,b])=[c,d]$ for some $c, d\in \mathbb{R}$.

Comments: Now, I'm pretty sure that this proof is fine, but the exercise is an old exam question which makes me think that the proof is overly convoluted and could be done much simpler. I'd like some help in that.

Answer 1

To prove that $f$ is continuous: let $U$ be an open interval in $[c,d]$. Then it is either of the form $(e,e')$ where $c<e<e'<d$, or of the form $[c,e)$, or of the form $(e,d]$. In the first case one proves that $f^{-1}(e,e')=(f^{-1}e,f^{-1}e')$ by showing that each side is contained in the other. The remaining cases are proved similarly.

The converse follows immediately from the fact that the image of a compact set (= closed and bounded in the case of $\mathbb{R}^n$) under a continuous map is compact.

Here I am using some facts from topology -- let me know if something is unclear.