In 3.11 of Rudin's Functional Analysis 2e:
If $X$ is infinite-dimensional then every weak neighborhood of $0$ contains an infinite-dimensional subspace; hence $X_w$ is not locally bounded.
Then it is sufficient to show that infinite-dimensional subspace is always unbounded, but there is no search result or in Wikipedia about this. Then what is the reasoning in the above content?
Suppose $X$ is an infinite dimensional topological space such that $X_w$ is locally bounded. Then, there are bounded linear functionals $\lambda_1,\ldots,\lambda_n$, and $\varepsilon>0$ such that $V=\{x\in X: |\lambda_j(x)|<\varepsilon,\,1\leq j\leq n\}$ is weakly bounded. $V$ contains contains the infinite dimensional subspace $Y=\bigcap^n_{j=1}\{x\in X:\lambda_j(x)=0\}$ and so, $Y$ is weakly bounded. Let $\lambda$ be a bounded linear functional that is not in the span of $\lambda_1,\ldots,\lambda_n$ (this is possible since $X$ is infinite dimensional). Since $Y$ is weakly bounded, there is $t_0>0$ such that $Y\subset t_0\{x\in X:|\lambda (x)|<1\}$. Thus $|\lambda(y)|<t_0$ for all $y\in Y$, and since $Y$ is a linear space, it follows that $\lambda(y)=0$ for all $y\in Y$. This implies that $\lambda\in\operatorname{span}(\lambda_1,\ldots,\lambda_n)$ which is a contradiction. Therefore, $X_w$ is not weakly bounded.