You are completely right, Eparoh, $\rm{span}\{\delta_k : k \in K\}$ is a subset of $C(K)^*$ that already separates points of $C(K)$.
I guess the issue is that Fabian et al. actually wanted state a stronger statement: Instead of considering the span of $\delta_k$ within $C(K)^*$, you could consider the $\delta_k$ as a "basis" and the vector space of finite linear combinations and prove the corresponding statement there.
More formally, let $V$ be the vector space of functions $g$ from $K$ to $\mathbb{R}$ such that $g(k)$ is non-zero only for a finite number of $k \in K$. Then $V$ corresponds to the finite linear combinations of elements of $\{g_k:k\in K\}$ were $g_k(k'):=1$ if $k=k'$ and $g_k(k'):=0$ otherwise.
We can map elements of $V$ to those of $\rm{span}\{\delta_k : k \in K\}$ via
the mapping $F:V\rightarrow \rm{span}\{\delta_k : k \in K\}$ by $F(\sum_{i=1}^n \alpha_i g_{k_i}) := \sum_{i=1}^n \alpha_i \delta_{k_i}$. So $g_k$ gets mapped to $\delta_k$. Define $\langle f,g\rangle := \langle f,F(g)\rangle$.
The intended statement is: $(C(K),V)$ is a dual pair.
To prove this we can identify $V$ with $\rm{span}\{\delta_k : k \in K\}$. For this we need to show that $F$ is injective (the surjectivity of $F$ is clear). Here, the Tietze-Urysohn theorem should be the way to go.
(Actually, this mean that the "wrong" formulation of Fabian et al just uses a kind of "equivalent classes" of linear combinations within $C(K)^*$ of deltas without caring that each of these classes consist of just one element. And Tietze-Urysohn theorem is used to see that these equivalent classes have just one element.)