 Solution

The intermediate value theorem states that, if $f$ is a continuous function on a closed interval $[a,b]$, every value between $f(a)$ and $f(b)$ equals $f(c)$ for some $c\in [a,b]$. If the value is strictly between the endpoint values, $c$ may be chosen in $(a,b)$.

Here is a proof from the supremum property of the real numbers. Suppose $f(a)<y<f(b)$; the reverse inequality is reduced to this case by replacing $f,y$ with $-f,-y$. LetThe set is nonempty because it contains $a$ and is bounded above by $b$. Continuity at $a$ supplies points of $E$ to the right of $a$, while continuity at $b$ excludes a neighborhood of $b$ from $E$. Hence $a<c<b$. By the definition of supremum, there are $t_n\in E$ with $t_n\to c$. Continuity gives $f(c)=\lim f(t_n)\le y$. If $f(c)<y$, continuity would give a point just to the right of $c$ still satisfying $f(t)<y$, contradicting that $c$ is an upper bound for $E$. Thus $f(c)=y$. Values equal to either endpoint are attained at that endpoint, completing the proof.