Taking $r < \frac{1}{2C}$ should work. Indeed, if a circle of radius $r$ touches the graph of $f$ by below at two points, then the curvature $$\kappa(x) = \frac{f''(x)}{(1+f'(x))^{3/2}}$$ is larger than $-\frac{1}{r}$ at the touching points. Since $f$ lies above the circle, $\kappa \leq -\frac{1}{r}$ somewhere in between the touching points, hence $f'' \leq -\frac{1}{r}$ there. Semiconvexity completes the proof.

Taking $r < \frac{1}{2C}$ should work. Indeed, if a circle of radius $r$ touches the graph of $f$ by below at two points, then the curvature $$\kappa(x) = \frac{f''(x)}{(1+(f'(x))^2)^{3/2}}$$ is larger than $-\frac{1}{r}$ at the touching points. Since $f$ lies above the circle, $\kappa \leq -\frac{1}{r}$ somewhere in between the touching points, hence $f'' \leq -\frac{1}{r}$ there. Semiconvexity completes the proof.

Taking $r < \frac{1}{2C}$ should work. Indeed, if a circle of radius $r$ touches the graph of $f$ by below at two points, then the curvature $$\kappa(x) = \frac{f''(x)}{(1+f'(x))^{3/2}}$$ is larger than $-\frac{1}{r}$ at the touching points. Since $f$ lies above the circle, $\kappa \leq -\frac{1}{r}$ somewhere in between the touching points, hence $f'' \leq -\frac{1}{r}$ there. Semiconvexity completes the proof.

A similar proof should work in higher dimensions, the idea being that if $D^2f geq -CI$ and we slide balls of radius less than $\frac{1}{2C}$ from below they must touch at only one point.

Taking $r < \frac{1}{2C}$ should work. Indeed, if a circle of radius $r$ touches the graph of $f$ by below at two points, then the curvature $$\kappa(x) = \frac{f''(x)}{(1+f'(x))^{3/2}}$$ is larger than $-\frac{1}{r}$ at the touching points. Since $f$ lies above the circle, $\kappa \leq -\frac{1}{r}$ somewhere in between the touching points, hence $f'' \leq -\frac{1}{r}$ there. Semiconvexity completes the proof.