= Solution
Let $C=C([0,1])$ with the uniform norm. For rational numbers $0\leq a<b\leq1$, let $M^+_{a,b}$ and $M^-_{a,b}$ be the sets of functions that are respectively nondecreasing and nonincreasing on $[a,b]$.
Both sets are closed. For example, if $f_j\in M^+_{a,b}$ and $f_j\to f$ uniformly, then for $x<y$ in $[a,b]$,
$$
f(x)=\lim_jf_j(x)\leq\lim_jf_j(y)=f(y).
$$
They also have empty interior. Given $f\in M^+_{a,b}$ and $\varepsilon>0$, continuity lets us choose $a<x<y<b$ sufficiently close that
$$
0\leq f(y)-f(x)<\frac{\varepsilon}{2}.
$$
Add a continuous triangular bump $h$ supported near $x$, with $h(x)=3\varepsilon/4$, $h(y)=0$, and $\lVert h\rVert_\infty<\varepsilon$. Then $(f+h)(x)>(f+h)(y)$, so the $\varepsilon$-ball about $f$ is not contained in $M^+_{a,b}$. The analogous upward bump at $y$ deals with $M^-_{a,b}$. Thus both sets are nowhere dense.
There are only countably many rational pairs $(a,b)$, so
$$
\bigcup_{a,b\in\mathbb Q,\ 0\leq a<b\leq1}
\left(M^+_{a,b}\cup M^-_{a,b}\right)
$$
is meagre. Completeness of $C$ and the <Baire category theorem> show that its complement is nonempty. Choose $f$ in that complement. If $f$ were monotone on an interval of positive length, that interval would contain a closed interval with rational endpoints, putting $f$ in one of the displayed sets. Hence $f$ is monotone on no interval of positive length, as described by the <generic nowhere-monotone continuous function> result.
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