Could someone please tell me where my mistake is in the following argument?
Suppose $V$ and $W$ are countably infinite dimensional vector spaces over a field $k$. Then $V=\varinjlim{V_i}$ and $W=\varinjlim{W_j}$, for some finite dimensional vector spaces $V_i$ and $W_j$. Now consider the dual space $(V \otimes W)^*$. We have $$(V \otimes W)^* =\operatorname{Hom}(V \otimes W, k) = \operatorname{Hom}(\varinjlim{V_i} \otimes \varinjlim{W_j}, k).$$ Since the tensor product commutes with direct limits, $$\varinjlim{V_i} \otimes \varinjlim{W_j}=\varinjlim (V_i \otimes \varinjlim{W_j})= \varinjlim \varinjlim (V_i \otimes W_j),$$ so we have $$\operatorname{Hom}(\varinjlim{V_i} \otimes \varinjlim{W_j}, k)= \varprojlim \varprojlim \operatorname{Hom}(V_i \otimes W_j, k).$$ But $V_i$ and $W_j$ are finite dimensional for all $i$ and $j$'s, and so $(V_i \otimes W_j)^*= V_i^* \otimes W_j^*$. This gives us $$\varprojlim \varprojlim \operatorname{Hom}(V_i \otimes W_j, k)= \varprojlim \varprojlim V^*_i \otimes W^*_j= (\varprojlim V^*_i) \otimes (\varprojlim W^*_j),$$ since both $V_i$ and $W_j$ are finite dimensional. But $\varprojlim{V_i^*} \cong V^*$ and $\varprojlim{W_j^*} \cong W^*$. Hence $(V \otimes W)^* \cong V^* \otimes W^*$.
We know that $(V \otimes W)^*$ is not necessarily $V^* \otimes W^*$ when $V$ and $W$ are infinite dimensional, so there must be a mistake somewhere in my proof, but where is it?