Prove that if $x,y\in X$ and $r\in\mathbb R$ are such that $f(x)<r<f(y), $ then $\exists z\in X:f(z)=r$

Question

Let $X$ be a connected topological space and $f:X\to\mathbb R$ a continuous function. Prove that if $x,y\in X$ and $r\in\mathbb R$ are such that $f(x)<r<f(y), $ then $\exists z\in X:f(z)=r.$

Proof. Consider $A=f(X)\bigcap (-\infty,r)$ and $B=f(X)\bigcap (r,\infty).$

Notice that $A$ and $B$ are nonempty, open and their intersection is empty.

Now assume there is no $z\in X$ such that $f(z)=r. $Then $f(X)=A\bigcup B$, i.e. $f(X)$ is not connected ! Therefore $\exists z\in X:f(z)=r.$

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Why the assumption of "no $z\in X$ such that $f(z)=r$" implies $f(X)=A\bigcup B$ ?

I think one can see $f(X)=A\bigcup B$ without such assumption: $A\bigcup B=(f(X)\bigcap (-\infty,r))\bigcup (f(X)\bigcap (r,\infty))=f(X)\bigcap ((-\infty,r)\cup(r,\infty))=f(X)\bigcap\mathbb R=f(X)$

Answer 1

Just to move this Q to the "answered" list:

$$(\neg \exists x\; (f(x)=r))\implies \emptyset= f(X)\cap \{r\}\implies$$ $$\implies f(X)=f(X)\cap \Bbb R=f(X)\;\cap\; [(-\infty,r)\,\cup \;\{r\}\;\cup\, (r,\infty)]=$$ $$=[f(X)\cap (-\infty,r)]\;\cup\; [f(X)\cap \{r\}]\;\cup \;[f(X)\cap (r,\infty)]=$$ $$=[A]\cup [\emptyset] \cup [B]=A\cup B.$$