= Solution
To prove the <intermediate value theorem> in this setting, let $A=\{t\in[a,b]:f(t)<v\}$ and $c=\sup A$. The endpoint inequalities and <continuity> give an interval immediately to the right of $a$ contained in $A$ and an interval immediately to the left of $b$ disjoint from $A$. Thus $a<c<b$. Choose points of $A$ tending to $c$; <continuity> gives $f(c)\le v$. If $f(c)<v$, <continuity> would place some point larger than $c$ in $A$, contradicting the supremum. Hence \b[$\boxed{f(c)=v}$] at an interior point.
For a continuous self-map of $[a,b]$, define $g(t)=f(t)-t$. The range condition implies $g(a)\ge0$ and $g(b)\le0$. If either is zero, its endpoint is a <fixed point>. Otherwise apply the just-proved <intermediate value theorem> to $-g$ with target zero to obtain $d\in(a,b)$ and $g(d)=0$. Therefore
$$
\boxed{\text{Every continuous self-map of }[a,b]\text{ has a fixed point.}}
$$
This proves the <fixed-point property of a closed interval> without requiring $f$ itself to be monotone.
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