I have done the following exercise , but I did not use one of the hypothesis so I wanted to make sure that everything was alright
Suppose that $X$ is a banach space then a subset $B$ is bounded if and only if $\forall f\in X' sup\{|f(b)|:b\in B\}< \infty$.
First suppose that $B$ is bounded so that for every $b\in B$ we will have that $||b||\leq M_B$, and so for any $f\in X'$ we have that $|f(b)|\leq ||f||||b||\leq ||f||M_B$ and so we can take the supremum and get the result.
Now for the other direction we consider the elements $J_b \in X''$ such that $J_b(f)=f(b)$ and $||J_b||=||b||$. Now the set $\{|f(b)|:b\in B\}$ is exactly $|J_b(f)|$ for fixed $f$ and we vary $b$ and by our hypothesis for every $f$ we will have that its supremum is finite and so since $X'$ is a Banach space we can use the Uniform Boundedness Principle to conclude that $sup_{b\in B} ||J_b|| <\infty$, and since $||J_b||=||b||$, we get the desired result. Now I did not use the fact that $X$ is a Banach space so I am wondering if I have made a mistake, any comment is aprecciated. Thanks in advance!
Completion is not required. If $X$ is not complete and $Y$ is the completion of $X$ then $\{F(b): b \in B\}$ is bounded for every $F \in Y'$. Hence, if the result holds in the complete case it holds for any normed linear space.