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.