In a the euclidean space $\mathbb{E}^n$ a line $\alpha$ is mapped to a line $\pi(\alpha)$ by an othogonal projection $\pi:\mathbb{E}^n\to P$ to some plane $P$. Projections in the hyperbolic space shares some properties with $\mathbb{E}$, for example projections to convex geodesic subspaces are contractive. Hence the following question.
Let then $\pi$ be the projection of $\mathbb{H}^n$ to some geodesic plane $P\subset H$: is it true that $\pi$ maps geodesic intervals $\ell$ to geodesic intervals $\pi(\ell)$?
Yes. The following proof works in $\mathbb{H}^3$ and is better understood if drawn while read, but I think it can be generalized to any dimension.
Recall that in the upper half-space model geodesic planes are vertical planes (i.e. orthogonal to the boundary) or hemispheres whose center lies on the boundary, while geodesic intervals are contained in straight lines orthogonal to the boundary or vertical semicircles (i.e. contained in a vertical plane) whose center lies on the boundary. From now on we will always assume that planes and circles are vertical.
We can assume without loss of generality that $P$ is a vertical plane. To project a point $x\notin P$ to $P$ take the unique circle $C=C(x,P)$ passing through $x$, orthogonal to $P$, and whose center lies in $P$, then the projection $\pi(x)$ is just the intersection $C\cap P$.
If the geodesic segment $\ell$ is contained in a vertical line then the statement is immediate, so suppose that $\ell$ is contained in a unique circle $C_\ell$. Let $S_\ell$ be the unique sphere that contains $C_\ell$ and whose center lies in $P$. Then for any point $x\in\ell$ the circle $C(x,P)$ projecting $x$ to $P$ is given by the intersection of this sphere with a plane orthogonal to $P$ and passing through $x$. This shows that $p(\ell)\subset S_\ell\cap P$ and hence conclude the proof, since it is contained in the circle.