b/ follows from a/, because . At any rate, that proves the triangle inequality. The other metric space axioms are straightforward.
The easiest way to prove the completeness of L(H) is to show first that .
Given a Cauchy sequence in L(H), we must first find a candidate for a limit operator T, and then show that in the operator norm (or more precisely, in the metric d associated with the norm).
So suppose that we have such a Cauchy sequence. For every v in H, it's not hard to see that is a Cauchy sequence in H. But H is complete, so this sequence converges to a vector that we call Tv. Check that the operator T defined in this way is linear.
To complete the proof, you have to show that (1) T is bounded, and (2) .
Let . For all sufficiently large n and m, . Therefore, for all v with , . In that inequality, you can let m→∞ to get . Taking the sup over all v in the unit ball, you get . That simultaneously establishes (1) and (2).