Let $S$ be the set of points of local minimum of $f$.

For each $x \in S$, by hypothesis, there exists an open interval $(\alpha_x, \beta_x)$ containing $x$ such that $f$ is strictly convex on $(\alpha_x, \beta_x)$.

Recall that a strictly convex function on an interval has at most one point of local minimum (since if there were two, the function would be constant or linear between them, contradicting strict convexity).

Therefore, for each $x \in S$, $x$ is the unique point of local minimum of $f$ in $(\alpha_x, \beta_x)$.

Thus, the intervals $(\alpha_x, \beta_x)$, for $x \in S$, are pairwise disjoint.

But the set of pairwise disjoint open intervals in $\mathbb{R}$ is at most countable (since each contains a rational number, and the rationals are countable).

Therefore, $S$ is at most countable.