Understanding Uniform Boundedness Theorem The following is from Bruckner's Real Analysis:

What does the green highlighted text mean at all?
The Baire Category Theorem states that a countable intersection of dense open sets is dense. So what are the dense sets here? Is the family $\mathcal{F}$ countable so to use BCT? BCT is about intersection of open sets not closed, but each ${\{x : \|Tx|| \le n}\}$ is closed? After these three questions be cleared explanation of the highlighted text would be much appreciated.
 A: By the Baire category theorem there exists $n\in\mathbb{N}$ such that $int(\overline{A_n})$ has non empty interior.
Fix this $A_n$.
Now use the fact that $A_n$ is $CS-closed$ (which means that the sum of each convergent convex series of element of $A_n$ is in $A_n$) which yields $int(A_n)=int(\overline{A_n})$.
Thus $\varnothing\neq int(A_n)$.
Then you observe that $A_n$ is convex and symmetric thus also $int(A_n)$ is convex and symmetric.
From this property conclude that $0\in int(A_n)$.
Hope this helps.
A: The reason is that Baire's theorem has a "dual form": if $\{A_n\}$ is a sequence of closed subsets of a Banach space $X$ with $Int(A_n)=\emptyset$ then $Int(\cup A_n)=\emptyset$. That is why you would be able to find a ball not intersecting $\cup A_n$. All of your $A_n$ have empty interior, otherwise your family would be uniformly bounded.
A: One simple way to understand:
We have that each $A_n$ is closed  and that $X = \bigcup_n A_n$.  So,  $X \setminus A_n$ is open and $\emptyset = \bigcap_n (X \setminus  A_n)$.
Now, by Baire Category Theorem (as you state it), if each $X \setminus A_n$ were dense, then $\bigcap_n (X \setminus  A_n)$ should be dense as well. Since  $\bigcap_n (X \setminus  A_n)= \emptyset$, we can conclude that, for at least one $n_0$, $X \setminus A_{n_0}$ is not dense. So, there is a ball $B(x_0, \delta)$ such that $B(x_0, \delta) \cap (X \setminus A_{n_0})=\emptyset$. It means $B(x_0, \delta) \subseteq A_{n_0}$.
Remark: The reasoning above is frequent when applying BCT and, in fact, it is completely general. So one equivalent formulation of BCT is:

A countable union of empty-interior closed sets has empty interior.

( empty-interior closed set is commonly called nowhere dense closed set. )
How to prove it? By taking complements, as I have done above.
