Sufficient condition for monotonicity

Take any $x_1<x_2$ in $[a,b]$. Since $f$ is continuous on $[x_1,x_2]$ and differentiable on $(x_1,x_2)$, Lagrange's Mean Value Theorem gives a point $c\in(x_1,x_2)$ with $f(x_2)-f(x_1)=f'(c)(x_2-x_1)$.

Because $f'(c)\ge 0$ and $x_2-x_1>0$, the right-hand side is $\ge 0$, so $f(x_2)\ge f(x_1)$. This already shows $f$ is non-decreasing on $[a,b]$.

To upgrade "non-decreasing" to strictly "increasing" despite the finitely many zeros of $f'$, list those zeros as $t_1<t_2<\dots<t_k$ inside $(x_1,x_2)$. On each open piece between consecutive points of the list $\{x_1,t_1,\dots,t_k,x_2\}$, $f'$ is strictly positive, so the same Mean Value Theorem argument applied to any two points inside that piece gives a strict inequality $f(u)<f(v)$ for $u<v$ there.

By continuity of $f$, the strict inequalities on consecutive pieces chain together: if $x_1<u<t_1<v<x_2$ say, then $f(x_1)<f(u)<f(t_1)\le f(v)<f(x_2)$, and since this holds for every choice of $u,v$ arbitrarily close to the endpoints, $f(x_1)<f(x_2)$ follows. Hence $f$ is (strictly) increasing on $[a,b]$. The decreasing case follows by applying this argument to $-f$.