Theorem: Let $f:X \to Y$ be a continuous map, where $X$ is a connected space and Y is an ordered set in the order topology. If $a,b \in X$ and $r \in Y$ such that $f(a) \lt r \lt f(b)$, then $\exists c \in X (f(c)=r)$
Proof Attempt: Let $f:X \to Y$ be a continuous map where $X$ is a connected space and $Y$ is an ordered set under the order topology. Suppose $a,b \in X$ and $r \in Y$ such that $f(a) \lt r \lt f(b)$. Since $X$ is connected then $f(X)$ is connected. Let $f(X)=(\gamma_0, \gamma)\cup (\overline{\gamma},\gamma')$. Since $f(a),f(b) \in f(X) \subset Y$, then $f(a),f(b)\in (\gamma_0, \gamma)\cup (\overline{\gamma},\gamma')$, then, w.l.o.g., let $f(b)\in (\overline{\gamma},\gamma')$, and $f(a)\in (\overline{\gamma},\gamma')$. Also, $f(X)$ is connected so $(\gamma_0,\gamma)\cap (\gamma,\gamma')\neq\emptyset$. This forces $r\in(\gamma_0,\gamma)\cup (\overline{\gamma},\gamma')=f(X)$. Therefore, $r\in f(X)$ and $\exists c\in X(f(c)=r)$.
Best Answer
Better: suppose we have $(Y,<)$ in the order topology and $f: X \to Y$ continuous with $X$ connected and for some points $a,b \in X$ we have $f(a) < r < f(b)$ with $r \in Y$.
Suppose there is no $x \in X$ with $f(x)=r$.
Then define $O_1=\{y \in Y: y < r \}$ and $O_2=\{y \in Y: y > r\}$ which are open in $Y$ in the order topology.
Firstly $$f^{-1}[O_1] \cup f^{-1}[O_2] = X$$
because we assumed $r$ is not assumed as a value, and $Y$ has a linear order, so always $f(x) > r$ or $f(x) < r$ must hold, but not both, so the sets are disjoint. We also know that $a \in f^{-1}[O_1]$ and $b \in f^{-1}[O_2]$, so both these sets are open (by continity of $f$), disjoint, non-empty and cover $X$. This is a contradiction with the connectedness of $X$. So we must have some $x$ with $f(x)=r$.