Suppose the latter is not the case, hence there is some x in H with Hx nonzero. By minimality, Hx = H. Find e in H with e*x = x.
Let B be the left anihilator of x, contained in H. Since B misses e, B = 0. Now e2-e kills x and lies in H, hence it lies in B, and e2-e = 0. This makes e an idempotent.
Since R*e is a nonzero submodule of H, it equals H, and H is a principal left ideal, generated by the idempotent e.
Together, R*e and R*(1-e) span R. If xe = y-ye, multiply by e on the right and xe = 0. The two left modules have no intersection, other than 0, and R is the direct sum of the two modules. In other words, H is a summand of R.
A division ring is semiprime, with 0 being a semiprime ideal. By correspondence, 0 is a semiprime ideal in the matrix ring, creating a larger semiprime ring. Take the direct product of semiprime rings, and R is semiprime. Thus semisimple implies semiprime and left (or right) dcc.
Conversely, let R be semiprime, with dcc on its principal left ideals. Given a left ideal H, let x generate a principal left ideal inside H. Keep extracting smaller principal left ideals; but this can't go on forever. It stops when e generates H, where H is a minimal left ideal.
If H2 = 0 then H = 0, since R is semiprime. Therefore, by the above lemma, we may assume e is idempotent, and H becomes a summand of R.
Since H is minimal it is a simple left R module. We have pulled a simple module off of R, and what remains contains another summand, and another, and another. At each step the complement, the part of R that remains, is generated by an idempotent, thus building a descending chain of principal left ideals. This cannot continue forever. Therefore R becomes a finite direct sum of simple left modules. This makes R a semisimple ring.
A similar proof equated semisimple with jacobson semisimple and principal left dcc, but this is more general. A jacobson semisimple ring has a lower nil radical of 0, whence 0 is a semiprime ideal, and the ring is semiprime. Yet there are semiprime rings that are not jacobson semisimple, so this is a stronger result.