A question about an $n$-dimensional subspace of $\mathbb{F}^{S}$.

For your basis $\mathcal B$, for each $x$ consider the $n$-dimensional vector with components $f_i(x)$. There is a linearly independent set of $n$ of these vectors. If $S$ is finite, this follows directly because the matrix formed by all these vectors has rank $n$ because $\mathcal B$ is a basis. It also holds for infinite $S$, however, for if not, some $n-1$ of these vectors would have to span all of them; the matrix formed by those $n-1$ vectors would have rank at most $n-1$, so it would be possible to express one of the functions as a linear combination of the others at the corresponding $n-1$ points, but then since the vectors corresponding to the remaining points are spanned by the $n-1$ vectors, the one function would in fact be identical to the linear combination of the others at all points, contrary to the fact that $\mathcal B$ is a basis.

So we have $n$ points $x_1,\dotsc,x_n$ such that the corresponding $n$ vectors $f_i(x_j)$ are linearly independent. But then their entries $A_{ij}=f_i(x_j)$ form an invertible matrix, and since

$$\delta_{ij}=\sum_kA^{-1}_{ik}A_{kj}=\sum_kA^{-1}_{ik}f_k(x_j)=\left(\sum_kA^{-1}_{ik}f_k\right)(x_j)$$

the points $x_1,\dotso,x_n\in S$ and the functions $\sum_kA^{-1}_{ij}f_k\in W$ have the desired property.

You can prove this by induction on $n$. Here's a sketch:

Base Case: ($n = 0$ is totally trivial). $n =1$: a one-dimensional subspace of $\mathbb{F}^S$ is the set of scalar multiples of a single not identically zero function $f: S \rightarrow \mathbb{F}$. So there exists some $x \in S$ such that $f(x) \neq 0$, and then by rescaling there exists some $x \in S$ and $\alpha \in \mathbb{F}$ such that $\alpha f(x) = 1$.

Inductive Step: Suppose that the result holds for any $n$-dimensional subspace $W = \langle f_1,\ldots,f_n \rangle$, and now suppose that we add to $W$ one linearly independent function $g$. By induction there is a subset $S_n = \{x_1,\ldots,x_n\}$ of $S$ such that that elements of $W$, when restricted to functions on $S_n$, give all possible functions on $S_n$. Therefore there is some linear combination of the $f_i$'s which induces the same function on $S_n$ as $g$ does, i.e., there are scalars $\alpha_1,\ldots,\alpha_n$ such that $(g - \sum_{i=1}^n \alpha_i f_i)(x_j) = 0$ for all $1 \leq j \leq n$. But since $g$ is linearly independent from $W$, $(g - \sum_{i=1}^n \alpha_i f_i)$ is not the zero function. Can you complete the argument from here?

By the way, I agree that double duality is also relevant. But I think the above approach is more "hands on" -- after proving it this way, one can think about what it means in terms of double dual spaces.

The solutions posted above can't be what H&K had in mind since they do not use the double dual. I came up with the following solution.

Let $s\in S$. We first show that the function \begin{alignat*}{1} \phi_s:&W\rightarrow F\\ &w\mapsto w(s) \end{alignat*} is a linear functional on $W$ (in other words for each $s$, we have $\phi_s\in W^*$).

Let $w_1,w_2\in W$, $c\in F$. Then $\phi_s(cw_1+w_2)=(cw_1+w_2)(s)$ which by definition equals $cw_1(s)+w_2(s)$ which equals $c\phi_s(w_1)+\phi_s(w_2)$. Thus $\phi_s$ is a linear functional on $W$.

Suppose $\phi_s(w)=0$ for all $s\in S$, $w\in W$. Then $w(s)=0$ $\forall$ $s\in S$, $w\in W$, which implies $\dim(W)=0$. So as long as $n>0$, $\exists$ $s_1\in S$ such that $\phi_{s_1}(w)\not=0$ for some $w\in W$. Equivalently there is an $s_1\in S$ and a $w_1\in W$ such that $w_1(s_1)\not=0$. This means $\phi_{s_1}\not=0$ as elements of $W^*$. It follows that $\langle\phi_{s_1}\rangle$, the subspace of $W^*$ generated by $\phi_{s_1}$, has dimension one. By scaling if necessary, we can further assume $w_1(s_1)=1$.

Now suppose $\forall$ $s\in S$ that we have $\phi_s\in\langle\phi_{s_1}\rangle$, the subspace of $W^*$ generated by $\phi_{s_1}$. Then for each $s\in S$ there is a $c(s)\in F$ such that $\phi_s=c(s)\phi_{s_1}$ in $W^*$. Then for each $s\in S$, $w(s)=c(s)w(s_1)$ for all $w\in W$. In particular $w_1(s)=c(s)$ (recall $w_1(s_1)=1$). Let $w\in W$. Let $b=w(s_1)$. Then $w(s)=c(s)w(s_1)=bw_1(s)$ $\forall$ $s\in S$. Notice that $b$ depends on $w$ but does not depend on $s$. Thus $w=bw_1$ as functions on $S$ where $b\in F$ is a fixed constant. Thus $w\in\langle w_1\rangle$, the subspace of $W$ generated by $w_1$. Since $w$ was arbitrary, it follows that $\dim(W)=1$. Thus as long as $\dim(W)\geq 2$ we can find $w_2\in W$ and $s_2\in S$ such that $\langle w_1,w_2\rangle$ (the subspace of $W$ generated by $w_1,w_2$) and $\langle\phi_{s_1},\phi_{s_2}\rangle$ (the subspace of $W^*$ generated by $\{\phi_{s_1},\phi_{s_2}\}$) both have dimension two. Let $W_0=\langle w_1,w_2\rangle$. Then we've shown that $\{\phi_{s_1},\phi_{s_2}\}$ is a basis for $W_0^*$. Therefore there's a dual basis $\{F_1,F_2\}\subseteq W_0^{**}$; so that $F_i(\phi_{s_j})=\delta_{ij}$, $i,j\in\{1,2\}$. By Theorem 17, $\exists$ corresponding $w_1,w_2\in W$ so that $F_i=L_{w_i}$ (in the notation of Theorem 17). Therefore, $\delta_{ij}=F_i(\phi_{s_j})=L_{w_i}(\phi_{s_j})=\phi_{s_j}(w_i)=w_i(s_j)$, for $i,j\in\{1,2\}$.

Now suppose $\forall$ $s\in S$ that we have $\phi_s\in\langle\phi_{s_1},\phi_{s_2}\rangle\subseteq W^{*}$. Then $\forall$ $s\in S$, there are constants $c_1(s),c_2(s)\in F$ and we have $w(s)=c_1(s)w(s_1)+c_2(s)w(s_2)$ for all $w\in W$. Similar to the argument in the previous paragraph, this implies $\dim(W)\leq 2$ (for $w\in W$ let $b_1=w(s_1)$ and $b_2=w(s_2)$ and argue as before). Therefore, as long as $\dim(W)\geq3$ we can find $s_3$ so that $\langle\phi_{s_1},\phi_{s_2},\phi_{s_3}\rangle\subseteq W^*$, the subspace of $W^*$ generated by $\phi_{s_1},\phi_{s_2},\phi_{s_3}$, has dimension three. And as before we can find $w_3\in W$ such that $w_i(s_j)=\delta_{ij}$, for $i,j\in\{1,2,3\}$.

Continuing in this way we can find $n$ elements $s_1,\dots,s_n\in S$ such that $\phi_{s_1},\dots,\phi_{s_n}$ are linearly independent in $W^{*}$ and corresponding elements $w_1,\dots,w_n\in W$ such that $w_i(s_j)=\delta_{i,j}$. Let $f_i=w_i$ and we are done.