Let $X$ be a normed vector space and $w_1,\ldots,w_n\in V$ be linearly independent. Assume $k\in\{1,\ldots,n\}$, $u_1,\ldots,u_k\in X$ with $\operatorname{span}\{u_1,\ldots,u_k\}=\operatorname{span}\{w_1,\ldots,w_k\}$ and $u_1^\ast,\ldots,u_k^\ast\in X'$ with $u_i^\ast(u_j)=\delta_{i,j}$ and $\left\|u_i\right\|_X=\left\|u_i^\ast\right\|=1$.
Can we construct $u_{k+1}\in X,u_{k+1}^\ast\in X'$ so that the former properties remain true?
We can clearly define $v_{k+1}:=w_{k+1}-\sum_{i=1}^ku_i^\ast(w_{k+1})u_i$ and $u_{k+1}:=\frac{v_{k+1}}{\left\|v_{k+1}\right\|}$. Then $U:=\operatorname{span}\{u_1,\ldots,u_{k+1}\}=\operatorname{span}\{w_1,\ldots,w_{k+1}\}$ and $u_i^\ast(u_{k+1})=0$ for all $i\le k$.
Now, we can clearly find $u_{k+1}^\ast\in U^\ast$ with $u_{k+1}^\ast(u_i)=\delta_{i,k+1}$ for all $i\le k$. Then by the Hahn-Banch theorem, $u_{k+1}$ has an extension to $X$ with the same operator norm.
The only missing property is $\left\|u_{k+1}^\ast\right\|=1$. Can we fix that?
Since Hahn-Banach does not increase the norm of the functional, it is enough to compute the norm of $u_i^*$ on the subspace $span(u_1\dots u_i)$: $$ \|u_i^*\| = \sup_{\|u\|\le1} u_i^*(u) =\sup_{\| \sum_{k=1}^i a_ku_k \|\le1} u_i^*(\sum_{k=1}^i a_ku_k) =\sup_{\| \sum_{k=1}^i a_ku_k \|\le1} a_i \ge \sup_{\sum_{k=1}^i |a_k|\le1} a_i =1. $$ The last equality is due to $\| \sum_{k=1}^i a_ku_k \| \le \sum_{k=1}^1i |a_k|$, hence last supremum is taken over a smaller set than the second-to-last.
The reverse inequality is a consequence of the dual basis property: $$ 1\le u_i^*(u_i) \le \|u_i^*\|\cdot \|u_i\| = \|u_i^*\|. $$