It suffices to show that under the given condition, $f$ is locally of bounded variation, because then $f$ can be written as a difference of two continuous non-decreasing functions, which are differentiable almost everywhere by the Lebesgue's theorem for the differentiability of monotone functions. So the following proposition gives the result.
Proposition: If $f$ is continuous on $[a,b]$ and takes each value at most $k$ times, then it holds
$$
V_a^b(f) \le k (M-m)
$$ where $V_a^b(f)$ is the total variation of $f$ on $[a,b]$, and $\displaystyle M = \max_{x\in [a,b]}f(x)$ and $\displaystyle m = \min_{x\in [a,b]} f(x)$.
Proof: Take any partition $\Pi = \{a=x_0<x_1<\cdots<x_{n-1}<x_n = b\}$. We denote $I_j = (f(x_{j-1}),f(x_{j}))$ if $f(x_{j-1})<f(x_{j})$, or $I_j = (f(x_{j}),f(x_{j-1}))$ otherwise. Define
$$
F(y) = \sum_{j=1}^n 1_{\{y\in I_j\}} = \big[\#\text{ of $j$ such that $y$ belongs to $I_j$}\big].
$$ Note that if $y\in I_j$, then by the Intermediate Value Theorem there exists $t_j \in (x_{j-1},x_j)$ such that $f(t_j)= y$. Since the number of such $t_j$ is at most $k$, this implies $F(y)\le k$ for each $y$. Because $I_j \subset [m,M]$ for each $j$, it follows that
\begin{align*}
\sum_{j=1}^n |f(x_j)-f(x_{j-1})| =& \sum_{j=1}^n \mathrm{len}(I_j)\\
=& \sum_{j=1}^n \int_{I_j}1\ \mathrm dy \\
=&\sum_{j=1}^n \int_m^M 1_{\{ y\in I_j\}} \mathrm dy\\
=& \int_m^M F(y) \mathrm dy \\
\le&\ k(M-m).
\end{align*} And this implies $$ V_a^b(f) = \sup_{\Pi} \sum_{j=1}^n |f(x_j)-f(x_{j-1})| \le k(M-m).$$
