As Wlodzimierz points out, the answer is trivially yes. Maybe it would help to recall that $S$ is compact, so that for each $x$ there exists $y$ such that $(x,y) \in S$ and $f(x) = y$. So if $f(x_1) = y_1$ and $f(x_2) = y_2$ then the point $\frac{1}{2}((x_1, y_1) + (x_2, y_2)) = (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$ belongs to $S$, and hence $f(\frac{x_1 + x_2}{2}) \geq \frac{y_1 + y_2}{2}$. Also it's easy to see that $f$ is continuous.

But it's interesting to note that the corresponding statement in ${\bf R}^3$ fails. Let $S$ be the convex hull of the circle in the $xy$-plane with center $(1,0,0)$ and radius $1$, and the point $(0,0,1)$. So $S$ is a cone. Now if we set $f(x,y) = {\rm max}_{(x,y,z) \in S} z$ for $(x,y)$ belonging to the disc with center $(1,0)$ and radius $1$, we find that $f(x,y) = 0$ on the boundary circle except at the point $(0,0)$, where it takes the value $1$.

As Wlodzimierz points out, the answer is trivially yes. Maybe it would help to recall that $S$ is compact, so that for each $x$ there exists $y$ such that $(x,y) \in S$ and $f(x) = y$. So if $f(x_1) = y_1$ and $f(x_2) = y_2$ then the point $\frac{1}{2}((x_1, y_1) + (x_2, y_2)) = (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$ belongs to $S$, and hence $f(\frac{x_1 + x_2}{2}) \geq \frac{y_1 + y_2}{2}$. Also it's easy to see that $f$ is continuous.

But it's interesting to note that continuity fails in ${\bf R}^3$. Let $S$ be the convex hull of the circle in the $xy$-plane with center $(1,0,0)$ and radius $1$, and the point $(0,0,1)$. So $S$ is a cone. Now if we set $f(x,y) = {\rm max}_{(x,y,z) \in S} z$ for $(x,y)$ belonging to the disc with center $(1,0)$ and radius $1$, we find that $f(x,y) = 0$ on the boundary circle except at the point $(0,0)$, where it takes the value $1$.