1. Understanding the Problem:
We are given a tetrahedron $ABCD$ and a point $X$ outside the tetrahedron such that the segment $XD$ intersects the face $ABC$ at an interior point. We need to prove that the sum of the lengths of the projections of $D$ onto the planes $XBC$, $XCA$, and $XAB$ is less than or equal to the sum of the distances from $D$ to $A$, $B$, and $C$.

2. Projections and Right Angles:
Let $A'$, $B'$, and $C'$ be the projections of $D$ onto the planes $XBC$, $XCA$, and $XAB$ respectively. By definition of projection, $\angle DA'C = \angle DB'C = \angle DC'A = 90^\circ$.

3. **Sphere of Diameter $DC$:**
Since $\angle DA'C = 90^\circ$, point $A'$ lies on the circle with diameter $DC$. Similarly, $B'$ and $C'$ lie on the circles with diameters $DA$ and $DB$ respectively.

4. Distance Inequality:
We need to show that $A'B' + B'C' + C'A' \leq DA + DB + DC$.

- Consider the projection $A'$ of $D$ onto the plane $XBC$. Since $A'$ lies on the circle with diameter $DC$, the distance $A'B'$ is at most $DC$.
- Similarly, $B'C'$ is at most $DA$ and $C'A'$ is at most $DB$.

5. Summing the Inequalities:
$$A'B' \leq DC$$
$$B'C' \leq DA$$
$$C'A' \leq DB$$

Adding these inequalities, we get:
$$A'B' + B'C' + C'A' \leq DC + DA + DB$$

6. Conclusion:
Therefore, we have shown that:
$$A'B' + B'C' + C'A' \leq DA + DB + DC$$

$\blacksquare$