Does weak compactness imply boundedness in a normed vector space (not necessarily complete)?

Does this argument work? Let $X$ be a normed vector space, with $A$ weakly compact in $X$. The collection of sets of the form

$$\{x \in X: |f(x) - f(a)| < 1 \},f \in X^*,a \in A$$

forms a cover of $A$ consisting of weakly open sets in $X$, and so should has a finite subcover.

For $x\in A$ define the map $T_x\colon X^*\to \Bbb K$ by $T_x(f)=f(x)$. Fix $f\in X^*$. Let for $n$ integer $O_n:=\{x\in X,|f(x)|<n\}$. These ones are weak open sets and form an open cover of $A$. So there is $N_f$ integer such that for all $x\in A$, $|T_x(f)|=|f(x)|<N_f$. As $X^*$ is complete, by the Principle of Uniform Boundedness, $\sup_{x\in A}\lVert T_x\rVert$ is bounded. By Hahn-Banach theorem, $\lVert T_x\rVert=\lVert x\rVert$, proving that $A$ is bounded.