One way of doing this (in $d=3,$ say) requires the "Archimedes axiom": if one convex body $K$ contains another body $L,$ then the perimeter of $K$ is greater than that of $L$ (the nicest proof uses Crofton's formula, which says that the perimeter is proportional to the measure of the set of lines which intersect the set). Then, you set $K$ to the hemisphere, and $L$ to your spherical convex polygon, and you are done. (Archimedes actually introduced this as an axiom in the Euclidean case when computing the perimeter of the circle, since he needed to know that the inscribed polygons provided a lower bound).
One way of doing this (in $d=3,$ say) requires the "Archimedes axiom": if one convex body $K$ contains another body $L,$ then the perimeter of $K$ is greater than that of $L$ (the nicest proof uses Crofton's formula, which says that the perimeter is proportional to the measure of the set of lines which intersect the set). Then, you set $K$ to the hemisphere, and $L$ to your spherical convex polygon, and you are done. (Archimedes actually introduced this as an axiom in the Euclidean case when computing the perimeter of the circle, since he needed to know that the inscribed polygons provided a lower bound).