The constant is $1/2$. Let $n=[d/d_0]\geq 1$, and let $t_k$ be the time which takes the long runner to arrive at the sistance $kd_0$ from the origin, $1\leq k\leq n$.
Proving by contradiction, suppose that on every interval $[(j-1)d_0,jd_0], j=1,...,k$ the average speed of the long distance runner is less than $v/2$. Then $t_n>2nd_0/v$. On the other hand the total time of the long distance runner is $d/v\geq t_n$. Therefore $$2nd_0/v<t_n\leq d/v\leq (n+1)d_0/v,$$ which implies $n<1$ a contradiction.
It is easily seen that $1/2$ is best possible. Let $d_0=d/2+\epsilon$ where $\epsilon>0$ is small. Let the long runner run first with very high speed half of the distance, then stop (or run very slowly), and then run with the same high speed to the end. The average speed on every interval of length $d_0$ is about $1/2$ of the overall average speed.
Same argument proves that $C\geq n/(n+1)$ when $n$ is known. Also notice that when $d$ is divisible by $d_0$, one can take $C=1$.