Here is an argument the avoids choice.
Suppose an element $\omega\in V^*\otimes W^*$ maps to $0$ in $(V\otimes W)^*$. Using the description of $V^*\otimes W^*$ as a quotient of the free vector space generated by symbols $\varphi\otimes \psi$, we can write $\omega=\sum_{i=1}^n \varphi_i\otimes \psi_i$, with $\varphi_i\in V^*$, $\psi_i\in W^*$. Let $V^f$, $W^f$ be the subspaces of $V^*$, $W^*$ spanned by $\varphi_i$, $\psi_i$ respectively. Because $V^f$ and $W^f$ are finite-dimensional, we can find bases $\alpha_1,\ldots,\alpha_n$ of $V^f$ and $\beta_1,\ldots,\beta_m$ of $W^f$. Let $\alpha:V\to K^n$ and $\beta:W\to K^m$ be the linear maps whose coordinate functions are the $\alpha_i$ and $\beta_i$, respectively. The maps $\alpha$ and $\beta$ are surjective because the $\varphi_i$ and $\psi_i$ are linearly independent.
Now consider the commutative diagram
$$
\begin{array}{ccc}
(K^n)^*\otimes (K^m)^*&\rightarrow&(K^n\otimes K^m)^*\\
\downarrow&&\downarrow\\
V^*\otimes W^*&\rightarrow&(V\otimes W)^*\\
\end{array}
$$
The element $\omega\in V^*\otimes W^*$ lifts to $(K^n)^*\otimes (K^n)^*$ by construction, the top horizontal map is an isomorphism because the spaces are finite dimensional, and map on the right is injective because $V\otimes W\to K^n\otimes K^m$ is surjective. So if $\omega$ maps to $0$ in $(V\otimes W)^*$, then the lift of $\omega$ in $(K^n)^*\otimes (K^m)^*$ must have been $0$, which implies $\omega=0$.
