Define $$g_n (x) := n \chi_{[0, \frac{1}{n}]}(x)$$ then $g_n\in L^1 (\mathbb R)\subset L^\infty(\mathbb{R})^*$, since $\|g_n\|_* = 1$, by weak-star compactness of the unit ball in $L^\infty(\mathbb{R})^*$, we know there exists a subsequence $$g_{n_k} \xrightarrow{\;*\;} \mu \in L^\infty(\mathbb{R})^*.$$
Now we also know that $g_n \rightarrow \delta_0$ in the sense of distributions. Is there any connection between $\mu$ and $\delta_0$? We know $\mu$ is a finite additive measure that is absolute continuous to the Lebesgue measure, thus it can not be $\delta_0$.
As was noted in comments, the unit ball of $(L^\infty)^*$ is not sequentially compact in the weak* topology. More specifically, your sequence $g_n$ does not have any weak*-convergent subsequence. Indeed, given any subsequence $g_{n_k}$, one can build an $L^\infty$ function $f$ (based on these indices $n_k$) such that $\int fg_{n_k}$ fails to have a limit. Here is a sketch:
(a) we may arrange that $n_{k+1}\ge 2n_k$ by extracting a further subsequence;
(b) let $f(x) = (-1)^k$ when $n_k<x^{-1}<n_{k+1}$;
(c) check that $\int fg_{n_k} \ge 1/3$ for even $n$ and $\int fg_{n_k} \le -1/3$ for odd $k$.
But it's true that the sequence $\{g_n\}$ has a weak* cluster point $\mu\in(L^\infty)^*$, meaning that every weak* neighborhood of $\mu$ is visited by the sequence infinitely often. (I find the cluster point language easier to digest than subnets.) Any such cluster point (there's no uniqueness) has the property that $$\mu(\phi) = \phi(0),\quad \forall \ \phi\in C(\mathbb R)$$ because $\int \phi g_n\to \phi(0)$ for continuous functions. Therefore: the restriction of any cluster point $\mu$ to the space of continuous functions agrees with the Dirac delta.
Informally speaking: continuous functions form a tiny slice of the incomprehensibly vast space $L^\infty$. The latter space has incomprehensibly many linear functionals that behave just like Dirac delta on continuous functions but act in different ways elsewhere in the space.