Answer
See proof below.
Work Step by Step
Let $s(t)$ be the function of his position, (s is in miles, t in hours) and let us assume that the function $s$ is continuous and differentiable over $[0,2.2].$
The function of velocity is the derivative of s, and we are given
$s'(0)=s'(2.2)=0$
We also assume that the runner does not change direction (he runs always towards the finish line), so we treat velocity as speed.
By the Mean Value Theorem, there exists a moment in time $t_{0}\in(0,2.2)$
where $s'(t_{0})=\displaystyle \frac{s(2.2)-s(0)}{2.2-0}=\frac{26.2}{2.2}=\frac{131}{11}\approx 11.09$ miles per hour
We have $s'(0)=0,\ s'(t_{0})=11.09,$
and, since $s'$ is continuous, by the Intermediate Value Theorem, it assumes all values between $0 $ and $11.09$ on $[0,t_{0}]$.
Thus, at least once in this interval, his speed was $11$ mph.
We also have $s'(t_{0})=11.09,\ s'(2.2)=0$,
So on the interval $[t_{0},2.2]$, his speed was $11$ mph at least once.
Thus, the runner's speed was $11$ mph at least twice.