I am trying to show that $U^* \otimes V \simeq \mathcal{L}(U,V)$ where $U$ and $V$ are vector spaces over $F$, $U^*$ is the dual space of $U$ and at least one of $U$ and $V$ is finite dimensional, but not necessarily both.
I used the universal property of tensor products and have shown that the map is injective but I have trouble showing it is surjective, since $U$ and $V$ are not both finite dimensional.
In both cases, the set $\mathcal L(U,V)$ becomes the set $\mathcal F(U,V)$ of the operators of finite rank. So we have to prove that there exists an isomorphism of vectorspaces $U^* \otimes V \simeq \mathcal F(U,V)$.
We start with a bilinear map $\beta \colon U^* \times V \to \mathcal F(U,V)$ with $$\beta(\varphi, v) \mapsto \varphi(\cdot)v.$$ This mapping is clearly bilinear and gives rise to a linear operator $T \colon U^* \otimes V \to \mathcal F(U,V)$.
You now should show that this operator is bijective.
Surjectivity: Let $f \in F(U,V)$ be given. The isomorphism theorem gives you an isomorphism $$\tilde f \colon {E}\,/\,{\ker f} \to \operatorname{range} f.$$ Let $[e_1],\dots,[e_n]$ be a basis in the quotient and let $[e_1]^*,\dots,[e_n]^*$ be their respective dual basis vectors. Now define functionals $$ e_i'(x) := [e_i]^*([x]). $$ Then, $$ f(x) = \tilde f([x]) = \tilde f\left(\sum_i [e_i]^*(x) [e_i]\right) = \left(\sum_i e_i'(\cdot) \,\tilde f([e_i]) \right)(x) = T\left( \sum_i e_i'\otimes f([e_i]) \right)(x), $$ which gives $$ f = T\left( \sum_i e_i'\otimes f([e_i]) \right). $$
Injectivity: Let $T\left(\sum_{i=1}^n \varphi_i \otimes v_i\right) = 0$. Then, $$T\left(\sum_{i=1}^n \varphi_i \otimes v_i\right)(x) =\sum_{i = 1}^n \varphi_i(x) v_i = 0$$ for all $x \in U$. By assuming $n$ to be minimal, we get that in particular $v_1,\dots,v_n$ are linearly independent which in turn gives that $\varphi_i(x) = 0$ for all $x \in U$. But then, $$\sum_{i =1}^n \varphi_i \otimes v_i = 0$$ and $T$ is injective.