Suppose $G = (V,E)$ is a finite, weakly connected, $d$-regular directed graph. Since $G$ is weakly connected, any cut that separates $V$ into two non-empty subsets is crossed by at least one edge.
The number of edges that cross said cut in each direction must be the same, or else $G$ wouldn't be finite and regular. In particular, there is at least one edge in each direction.
If there are vertices $u$ and $v$ such that there is no path in $G$ from $u$ to $v$, there must be a cut in $G$ such that $u$ and $v$ are on opposite sides of the cut and there is no edge of $G$ crossing from $u$'s side to $v$'s side.
But $d$-regularity means that there is no edge across that cut in the opposite direction either, contradicting the weak connectedness of $G$.