Prove that the uniform boundedness principle can be derived from the closed graph theorem. Let $(X,\|\cdot \|_X)$ and $(Y,\|\cdot \|_Y)$ be real Banach spaces. Let $\{ T_\alpha\}_{\alpha\in \Delta}\subseteq B(X,Y)$, where $B(X,Y)$ is the space of all bounded linear maps. Define a new norm on $X$ by
$$\|x\|'=\|x\|_X+\sup\limits_{\alpha\in \Delta}\|T_{\alpha}x\|_Y.$$
Prove that the uniform boundedness principle can be derived from the closed graph theorem.
My trial
By assumption, 


*

*$\{ T_\alpha\}_{\alpha\in \Delta}\subseteq B(X,Y)$

*$\sup\limits_{\alpha\in \Delta}\|T_{\alpha}x\|_Y=\|x\|'-\|x\|_X<\infty.$
Let ${\alpha\in \Delta}$, define \begin{align} G(T_\alpha)=\big\{(x,T_\alpha x):x\in X  \big\} \subseteq X\times Y  .\end{align}
Clearly, $G(T_\alpha)$ is a closed subset of $X\times Y$ which is Banach. Define \begin{align} \Pi_1:&\left(G(T_\alpha), \|\cdot \|_{X\times Y}\right)\to (X,\|\cdot \|_X)\\&(x,y)\mapsto  \Pi_1(x,y)=x . \end{align}
Clearly, $\Pi_1$ is bijective, linear and bounded. So, $\Pi_1^{-1}$ is continuous, and there exists $M\geq 0$ such that 
$$\|\Pi_1^{-1}(x) \|_{X\times Y}\leq M\|x\|_X.$$
This means that 
$$\|x\|_X+\sup\limits_{\alpha\in \Delta}\|T_{\alpha}x\|_Y=\|(x,y) \|_{X\times Y}\leq M\|x\|_X.$$
This results to 
$$\sup\limits_{\alpha\in \Delta}\|T_{\alpha}\|_Y\leq M.$$
Please, I'm I right? If no, then alternative proofs are highly welcome.
 A: Notice that the $\Pi_1$ you define also depends on $\alpha$ and hence so does the constant $M$. In particular, you cannot just take the $\sup$ to conclude that
$$\|x\|_X+\sup\limits_{\alpha\in \Delta}\|T_{\alpha}x\|_Y=\|(x,y) \|_{X\times Y}\leq M\|x\|_X.$$
Instead, consider the space $Z = \{f: \Delta \to Y \mid \sup_{\alpha \in \Delta} \|f(\alpha)\|_Y < \infty\}$ of bounded functions from $\Delta$ to $Y$ with the $\sup$-norm $\|f\|_\infty = \sup_{\alpha \in \Delta} \|f(\alpha)\|_Y$. It is a standard exercise to check that $Z$ is a Banach space (similar to checking that the space of continuous real-valued functions is a Banach space for the $\sup$-norm). 
Define $\Psi:X \to Z$ by $\Psi(x)(\alpha) = T_\alpha x$. By assumption, this map is well-defined and is clearly linear. If $(x_n, \Psi(x_n)) \to (x,f)$ in $X \times Z$, then for $\alpha \in \Delta$,
$$f(\alpha) = \lim_{n \to \infty} \Psi(x_n)(\alpha) = \lim_{n \to \infty} T_\alpha x_n = T_\alpha x$$ since each $T_\alpha$ is bounded. Hence $f = \Psi(x)$ and so $\Psi$ has closed graph and hence is bounded by the closed graph theorem. 
Finally, for each $\alpha$,
$$ \|T_\alpha x\|_Y = \|\Psi(x)(\alpha)\|_Y \leq \|\Psi(x)\|_\infty \leq \| \Psi \| \cdot \|x\|_X$$
and so $\sup_{\alpha \in \Delta} \|T_\alpha\| \leq \|\Psi\| < \infty$ (and, in fact, one can check easily that $\sup_{\alpha \in \Delta} \|T_\alpha\| = \|\Psi\|$).
