About uniform boundedness theorem. (Uniform Boundedness Thm) $X : $ Banach space, $Y$ : Normed vector space, $\mathcal A \subset B(X,Y) := \{ f : X \to Y \; | \; f : \text{bounded operator} \} $. If for any $x \in X$, $\sup_{T \in \mathcal A} \| Tx \|_Y < \infty $, then $\sup_{T \in \mathcal A} \| T \| < \infty.$
I am studying this theorem, but in the proof, I could not know that why $F_n := \{ x \in X \; | \; \sup_{T \in \mathcal A} \| Tx \|_Y \leqslant n \} $ is closed for all $n$, and $X = \bigcup_{n=1}^\infty F_n$ . 
 A: This is extended version of Qiaochu's comment
1) Note that
$$
x\in F_n\Longleftrightarrow
\sup_{T\in\mathcal{A}}\Vert Tx\Vert\leq n \Longleftrightarrow
\forall T\in\mathcal{A}\quad\Vert Tx\Vert\leq n \Longleftrightarrow\\
\forall T\in\mathcal{A}\quad Tx\in B_Y(0,n) \Longleftrightarrow
\forall T\in\mathcal{A}\quad x\in T^{-1}(B_Y(0,n)) \Longleftrightarrow\\
x\in\bigcap_{T\in\mathcal{A}}T^{-1}(B_Y(0,n))
$$
and we conclude
$$
F_n=\bigcap_{T\in\mathcal{A}}T^{-1}(B_Y(0,n))\tag{1}
$$
The ball $B_Y(0,n)$ is a closed set. Since $T$ is continuous, then preimage $T^{-1}(B_Y(0,n))$ of closed set is closed. Intersection of closed sets is closed, so from $(1)$ we conclude that $F_n$ is closed.
2) Obviously 
$$
\bigcup\limits_{n\in\mathbb{N}} F_n\subset X\tag{2}
$$
Take arbitrary $x\in X$ and consider natural number $N=\lfloor \sup_{T\in\mathcal{A}}\Vert Tx\Vert\rfloor+1$. Then $x\in F_N\subset \bigcup\limits_{n\in\mathbb{N}} F_n$. Since $x\in X$ is arbitrary we see that
$$
X\subset \bigcup\limits_{n\in\mathbb{N}} F_n\tag{3}
$$
From $(2)$ and $(3)$ the desired equality follows.
