If $U,V$ are bounded and closed intervals and $f:U \rightarrow \mathbb R$ is continuous, then $V \subset f(U)$ entails the existence of an interval $U_0 \subset U$ such that $f(U_0)=V.$
In my book, this lemma is stated but not proved and I cannot figure out how to prove it by myself.
The set $V$ is a bounded and closed interval, so it has a maximal and minimal value, say $V=[x,y]$ (I assume $x\ne y$, since this case is trivial). Then consider the sets $X=f^{-1}(x)$ and $Y=f^{-1}(y)$ which both, as closed subsets of the bounded and closed, hence compact(by Heine-Borel), interval $U$, are compact.
But for two compact sets the minimal distance is attained, so $d(X,Y)=\epsilon >0$ since $d(X,Y)=0$ would contradict continuity of $f$.
Since the minimal distance between $X$ and $Y$ is attained we can pick $a\in X$ and $b\in Y$, wlog $a