Let $q$ be a prime power and let $P$ be a projective plane of order $q$. It has $q^2+q+1$ points and $q^2+q+1$ lines. Each point lies on $q+1$ lines, and each line has $q+1$ points. Each pair of lines has exactly 1 common point. Since the point-line incidence graph is regular and bipartite, there is a bijection $L$ from points to lines such that $L(v)$ is a line through $v$ for every point $v$.

Now construct a graph $G$. The vertex set $V$ is the set of points of $P$. The out-neighbourhood of a vertex $v$ is $L(v)-v$, i.e. $e_v=L(v)$. So $n=q^2+q+1$ and $d=q$.

Now consider distinct $v'_1,\ldots,v'_{q+1}$. Since distinct lines have one common point, $|L(v'_1)\cup\cdots\cup L(v'_t)|\ge \sum_{i=1}^{q+1} (q-i+2)=\frac12(q+2)(q+3)>\frac12n+2q$. Thus, after $q+1=O(n^{1/2})$ steps, already less than half the vertices are available for the $\{v_i\}$ sequence. Even if all the remaining vertices can be chosen (most unlikely), in total less than half the vertices can be chosen.

I suspect the last part of this argument is unnecessarily weak and that the real bound is a lot smaller than $n/2$.

This example doesn't strictly violate the conjecture as stated since $C$ is allowed to be a function of $d$ which is a function of $n$.

Let $q$ be a prime power and let $P$ be a projective plane of order $q$. It has $q^2+q+1$ points and $q^2+q+1$ lines. Each point lies on $q+1$ lines, and each line has $q+1$ points. Each pair of lines has exactly 1 common point. Since the point-line incidence graph is regular and bipartite, there is a bijection $L$ from points to lines such that $L(v)$ is a line through $v$ for every point $v$.

Now construct a graph $G$. The vertex set $V$ is the set of points of $P$. The out-neighbourhood of a vertex $v$ is $L(v)-v$, i.e. $e_v=L(v)$. So $n=q^2+q+1$ and $k=q$.

Now consider distinct $v'_1,\ldots,v'_{q+1}$. Since distinct lines have one common point, $|L(v'_1)\cup\cdots\cup L(v'_t)|\ge \sum_{i=1}^{q+1} (q-i+2)=\frac12(q+2)(q+3)>\frac12n+2q$. Thus, after $q+1=O(n^{1/2})$ steps, already less than half the vertices are available for the $\{v_i\}$ sequence. Even if all the remaining vertices can be chosen (most unlikely), in total less than half the vertices can be chosen.

I suspect the last part of this argument is unnecessarily weak and that the real bound is a lot smaller than $n/2$.

This example doesn't strictly violate the conjecture as stated since $C$ is allowed to be a function of $(k,d)$ which is a function of $n$.