= Solution
The <intermediate value theorem> says that if $f:[a,b]\to\mathbb R$ is a <continuous function>, then every value between $f(a)$ and $f(b)$ is attained at some point of $[a,b]$.
Endpoint values are already attained. Suppose $f(a)<y<f(b)$ and define $E=\{x\in[a,b]:f(x)<y\}$. It is nonempty and bounded above, so $c=\sup E$ exists. Continuity at $a$ ensures that $E$ contains points to the right of $a$, and continuity at $b$ ensures that no point sufficiently close to $b$ belongs to $E$. Hence $a<c<b$.
By the definition of the <supremum>, there are points of $E$ approaching $c$. If $f(c)>y$, continuity would exclude all such nearby points from $E$, a contradiction. If $f(c)<y$, continuity would put some point to the right of $c$ in $E$, again a contradiction. Therefore \b[$f(c)=y$]. When $f(a)>f(b)$, apply this argument to $-f$. This proves the <intermediate value theorem> in all cases.
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