Let $f_1,\ldots,f_m \in V^*$ linearly independent. Show that each $1\leq j \leq m$ exists ($v_j$) in $V$ such that $f_i(v_j)=\delta_{ij}$.
I'm using the bidual space and evaluation function to prove the theorem in the case that $\operatorname{Dim}(V)$ is finite. But the case where $\operatorname{Dim}(V)$ is infinite I can't prove it.
Let $N$ denote the span of $\{f_1, \dots, f_m\} \subseteq V^*$. Consider the linear map $\varphi: V \to N^*$ defined by $\varphi(v)(f) = f(v)$ for $v \in V$ and $f \in N$. Then $\rm{image}(\varphi) \subseteq N^*$, evidently. On the other hand, show $N$ imbeds in $\rm{image}(\varphi)^*$. Conclude that $N$ and $\rm{image}(\varphi)$ have the same (finite) dimension, and hence that $\varphi$ is surjective. You have now essentially reduced to the finite dimensional case.