$\textbf{Banach-Alaoglu Theorem}$ states that if $X$ be a separable Banach space. Then every bounded sequence of linear functionals $\psi_n \in X^\ast$ admits a weak-star convergent subsequence.
The theorem is proved as the following steps;
$1)$ Since X is separable, there exists a dense countable set $S$.There exists a subsequence $(\psi_{n_j} )_{j\ge 1}$ which converges pointwise at each point $x_k \in S$, so that $\lim_{j\to \infty}\psi_{n_j}(x_k) = \psi (x_k)$, all $k \ge 1$.
$2)$ The limit function $\psi : S \to \mathbb{K}$ is Lipschitz continuous with constant $C = \sup_n \|\psi_n\|_\ast$. Therefore, the map $\psi$ can be uniquely extended by continuity to the closure of $S$, i.e., to the entire space $X$.
$3)$ Subsequence $\psi_{n_j}$ weak-star converges to $\psi$.
$\textbf{Question:}$
I have problem to understand the intuition behind the second step.Why do we need liptschitz continuity?Can't we proceed by proving $\psi$ is bounded? Where did we use the Completeness of $X$?
In part $(1)$ , it was proved that we have a pointwise convergent (sub)sequence on $S$.
In part $(2)$, it was proved the $\psi$ is Lipschitz continuous, hence bounded and using Bounded Extension (Hahn-Banach) Theorem we can extend $\psi$ to $X$.
In part $(3)$, then showing that $\psi_{n_j} \rightharpoonup^\ast \psi$.
The limit function $\psi$ is proved bounded by the Lipschitz condition holding. It is perfectly possible for a bounded sequence of functionals to converge to an unbounded functional -- if it weren't, then you wouldn't need the second step at all as we'd already know that $\psi$ was bounded and hence continuous. Constructing such a sequence is an easy exercise :)
Without completeness you cannot know that the closure of the dense subset $S$ is $X$. It is possible that $X \subset \bar{S}$ is a proper inclusion (for example, if your $S$ is a proper dense subset of the open unit ball in $\mathbb{R}^2$, the closure of $S$ will be the closed unit ball in $\mathbb{R}^2$ which is strictly larger than the open unit ball).