You can prove that compacts in the weak topology are bounded without the Eberlein-Šmulian theorem?
With Smuliam's Theorem this is immediate, because if you assume by contradiction that $K$ is unborded in the norm topology, exists $\{x_n\}_k \subset K$ such that $||x_k||>k$ for all $k$, due to the compactness in the weak topology, being relatively compact, there is a limit point, so the sequence cannot explode in norm. I would like to know if there is any way to do this without using this Theorem?
Correct me if I'm missing something subtle, but I think this follows directly from the uniform boundedness principle. Let $X$ be a Banach space and let $K \subset X$ be weakly compact. Consider $K$ as a subset of $X^{**}$. Each $f \in X^*$ is continuous with respect to the weak topology on $X$, so the set $\{f(x) : x \in K\} = f(K)$ is a compact subset of $\mathbb{R}$ or $\mathbb{C}$ and hence bounded. Thus the set $K \subset X^{**}$ is pointwise bounded, and so by the uniform boundedness principle it is bounded in norm. Since the norm on $X^{**}$ agrees with the norm on $X$ (the canonical embedding is an isometry) we have shown that $K$ is bounded in norm.