Let $V$ be a vector space over a field $F$, and let $\varphi_1,\ldots,\varphi_k \in V^*$, be linearly independent.
Claim: If $V$ is finite-dimensional, then there exist $x_j \in V$ such that $\varphi_i(x_j) = \delta_{ij}$. In other words: the $\varphi_i$ are the dual set of the $x_i$.
Question: Does this claim hold when $V$ is infinite-dimensional?
Edit:
If $V$ is finite-dimensional, and we fix a non-degenerate bilinear form $\langle,\,\rangle$ on $V$, then every $\varphi_i$ is given by $x \mapsto \langle x,u_i\,\rangle$, and we are looking for $x_j$ satisfying $\langle x_j,u_i\,\rangle = \delta_{ij}$.
This problem looks similar (but non-identical) to the Gram–Schmidt process.
Here is a proof for the finite-dimensional case:
Let $\alpha_1,\ldots,\alpha_k \in (V^*)^*$ be the dual set to $\varphi_i$, i.e. $\alpha_j(\varphi_i)=\delta_{ij}$.
Since $\dim V< \infty$, we have $(V^*)^* \cong V$ by the evaluation map, hence for every $\alpha_j \in (V^*)^*$ there exist $x_j \in V$ satisfying $$ \alpha_j(\varphi )=\varphi(x_j) $$ and in particular $$ \delta_{ij}=\alpha_j(\varphi_i )=\varphi_i(x_j). $$
While the proof uses the isomorphism $V \cong V^{**}$, which does not hold when $\dim V= \infty$,a-priori, there could be an alternative argument for the existence of the $x_j$ which does not rely on this isomorphism. At least I couldn't rule out this possibility.
I think it can be done by simple induction.
If $k = 1$ then there is $x_1'$ s.t. $\phi(x_1') \neq 0$, and then $x_1 = x_1' / \phi(x_1')$ will work.
Assume $\phi_1, \ldots, \phi_{k - 1}$ is dual set of $x_1, \ldots, x_{k - 1}$.
Let $\phi_k'(x) = \phi_k(x - \phi_1(x)\cdot x_1 - \ldots \phi_{k - 1}(x)\cdot x_{k - 1})$.
If $\phi_k'(x)$ is always zero, then $\phi_k = \sum\limits_{i=1}^{k - 1} \phi_i \cdot \phi_k(x_i)$, and thus our system isn't independent.
Otherwise, let $x'_k$ be such that $\phi_k'(x_k') \neq 0$. Let $x''_k = x'_k - \sum\limits_{i=1}^{k - 1}\phi(x'_k)\cdot x_i$.
From definition, $\phi_k(x''_k) \neq 0$ and it's easy to check that $\phi_i(x''_k) = 0$ if $i < k$.
Finally, let $x_k = x''_k / \phi_k(x''_k)$.
Now, $x_1 - \phi_k(x_1) \cdot x_k, x_2 - \phi_k(x_2) \cdot x_k, \ldots, x_k$ will be dual set for $\phi_1, \ldots, \phi_k$.