Suppose that T is the operator T(x₁, x₂, ¼) = (x₂, x₃, ¼) on l² ( in fact T is the adjoint of the unilateral shift operator). If T has a square root S, then S² = T and KerS Í KerT = C x₁ in which x₁ = (1, 0, 0, ¼). Since T is not one to one we conclude that S is not one to one. So that KerS = C x₁.  T is surjective, hence S is onto. So there exists an element h such that Sh = x₁. Since Th = S²h = Sx₁ = 0, we have h = lx₁ for some l Î C and hence x₁ = Sh = lS x₁ = 0, a contradiction.

Comment. There is an open subset of L(H) consisting of invertible operators with no square roots.
Ref.
J.B. Conway and B.B. Morrel, Roots and logarithms of bounded operators on Hilbert spaces, J. Funct Anal, 70(1987), 171-193.