it will suffice to exhibit a basis for V, and then show that for at least one element v of the basis (u,v)H ≠ 0.

the natural basis that comes to mind is: {(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)}.

let u = (u1,u2,u3,u4).

then (u,(1,0,0,0))H = u1(1^2) + u2(0^2) + u3(0^2) + u4(0^2) = u1

similarly, (u,(0,1,0,0))H = u2, (u,(0,0,1,0))H = u3, (u,(0,0,0,1))H = u4.

since by assumption, u is non-zero, one of u1,u2,u3 or u4 is non-zero, so one of those 4 inner products with basis vectors for V is non-zero.

in my opinion, it's far more interesting to show (u,v)H is conjugate-linear in v, and conjugate symmetric (which then implies (u,v)H is positive-definite), linearity in u is obvious.