As P and S are the midpoints of AB and AC respectively, the ratio AP/AB = AS/AC = 1/2.
This, along with the fact that triangles ABC and APS share angle BAC is sufficient to say that APS and ABC are similar triangles, so we can conclude that PS is parallel to BC.
Similar logic in triangle BCS allows us to conclude that QR is also parallel to BC, and therefore PS is parallel to QR.
Repeating the argument with the other 2 faces tells us that the remaining sides are also parallel to each other, so PQRS is a parallelogram.
If the tetrahedron is regular and of side s, the similar triangles show us that PQ = QR = RS = PS = 0.5s. So PQRS is a rhombus. By symmetry we can see that all angles of the rhombus must be equal, so we in fact have a square.
(You can use a similar technique to prove that the quadrilateral formed by joining all 4 midpoints of any quadrilateral is also a parallelogram)