proof of closed graph theorem

Let $T:X\to Y$ be a linear mapping. Denote its graph by $G(T)$, and let $p_1:X\times Y\to X$ and $p_2:X\times Y\to Y$ be the projections onto $X$ and $Y$, respectively. We remark that these projections are continuous, by definition of the product of Banach spaces.

If $T$ is bounded, then given a sequence$\{(x_i,T{x}_i)\}$ in $G(T)$ which converges to $(x,y)\in X\times Y$, we have that

$$x_i=p_1(x_i,T{x}_i)\underset{i\to \mathrm{\infty }}{\overset{}{\to }}{p}_1(x,y)=x$$

and

$$T{x}_i=p_2(x_i,T{x}_i)\underset{i\to \mathrm{\infty }}{\overset{}{\to }}{p}_2(x,y)=y,$$

by continuity of the projections.But then, since $T$ is continuous,

$$Tx=\underset{i\to \mathrm{\infty }}{lim}T{x}_i=y.$$

Thus $(x,y)=(x,Tx)\in G(T)$, proving that $G(T)$ is closed.

Now suppose $G(T)$ is closed. We remark that $G(T)$ is a vector subspace of $X\times Y$, and being closed, it is a Banach space. Consider the operator$\tilde{T}:X\to G(T)$ defined by $\tilde{T}x=(x,Tx)$. It is clear that $\tilde{T}$ is a bijection, its inverse being ${p_1|}_{G(T)}$, the restriction of $p_1$ to $G(T)$. Since $p_1$ is continuous on $X\times Y$, the restriction is continuous as well; and since it is also surjective, the open mapping theorem implies that ${p_1|}_{G(T)}$ is an open mapping, so its inverse must be continuous. That is, $\tilde{T}$ is continuous, and consequently $T=p_2\circ \tilde{T}$ is continuous.