I am reading a book in which a theorem is only proven analogously in the situation of weak topology. The proof does not seem to work in the case of weak$^{*}$ topology.
Let $X$ be a normed linear space. By definition, the weak$^{*}$ topology of $X^{*}$ is generated by the collection of seminorms $\{q_x : x \in X\}$, where $q_x (f) = |f(x)|$.
A theorem says that
A linear functional on $X^{*}$ is weak$^{*}$ continuous if and only if it belongs to $K= \iota(X)$, with the embedding $\iota:X \rightarrow X^{**}$ defined by $(\iota(x))(f)= f(x)$.
The "if" direction is easy. For the "only if" direction, by standard results, there exist $x_1 , \ldots, x_n \in X$ and $\alpha >0$ such that $$ |g(f) | \leq \alpha \sum_{i=1}^{n} q_{x_i} (f) := \alpha \sum_{i=1}^n |f(x_i)|, \quad \quad \forall f \in X^{*}.$$
This time, unlike in the case of weak topology, the kernel argument fails. How can we show that $g= \iota (x)$, for some $x \in X$?
Here is the idea.
Suppose there are functionals $g_1,...,g_n$ and $g$ such that $\cap_k \ker g_k \subset g$ then we can write $g = \sum_k \eta_k g_k $ for some $\eta_k$.
To see this, let $\phi:X \to \mathbb{C}^n$ be defined by $\phi(x) = (g_1(x),...,g_n(x))$ and note that $\phi(x) = \phi(y) $ means that $g(x) = g(y)$. In particular, if we define the equivalence $x \sim y$ iff $\phi(x-y) = 0$, then we see that $\tilde{\phi} : X / \sim \, \to \mathbb{C}^n$ defined by $\tilde{\phi} ([x]) = \phi(x)$ is well defined and an isomorphism and $\tilde{g} ([x]) = g(x)$ is well defined. Let $h: \mathbb{C}^n \to \mathbb{C}$ be $h = \tilde{g}\circ \tilde{\phi}^{-1}$,we see that $h$ is linear, hence $h(z) = \eta^T z$ for some $\eta$, and so $g(x) = \tilde{g}([x]) = (h \circ \tilde{\phi})([x])= (h \circ \phi)(x) = \sum_k \eta_k g_k(x)$, or, in other words, $g = \sum_k \eta_k g_k$.
Now note that the last inequality in the question implies a containment of the intersection of kernels and hence we can write $g = \sum_k \alpha_k \iota(x_k)$ for some $\alpha_k$.