Tensor products of infinite-dimensional spaces and other objects
No. As Theo says in the comments, the elements of $U^{\ast} \otimes V$ are precisely the finite-rank maps in $\text{Hom}(U, V)$.
This isn't even true in finite dimensions. The dimension of the LHS grows linearly in $\dim A$ but the dimension of the RHS grows quadratically in $\dim A$.
This also isn't even true in finite dimensions. The dimension of the LHS grows quadratically in $\dim C$ but the dimension of the RHS grows linearly in $\dim C$.
You seem to be under the mistaken impression that the tensor product is supposed to behave like a product. It isn't. Abstractly it comes from the tensor-hom adjunction
$$\text{Hom}(A \otimes B, C) \cong \text{Hom}(A, \text{Hom}(B, C))$$
and concretely it comes from wanting the free vector space functor $\text{Set} \to \text{Vect}$ to be (lax?) monoidal.
Working with naked infinite-dimensional vector spaces is asking for trouble. See topological tensor product for appropriate substitutes for topological vector spaces.