given u in U, it is enough to show that φ(u) is in Ann(Ann(U)) (this proof only works in finite-dimensional vector spaces, by the way, as in general, U is not isomorphic to U**).
but if f is in Ann(U), this means f(u) = 0 for all u in U. hence for any such f, φ(u)(f) = f(u) = 0, that is φ(u) is contained in Ann(Ann(U)), which means that φ(U) = U** is contained in Ann(Ann(U)).
but clearly Ann(Ann(U)) is a subset of U**, so the two sets are equal (the finite-dimensionality comes in by way of asserting φ is ONTO U**).
now suppose we take G in Ann(Ann(U)). since Ann(Ann(U)) = U**, and φ is an isomorphism, φ-1 is well-defined. letting u = φ-1(G), we have:
φ(u) = φ(φ-1(G)) = idU**(G) = G.