Let $T\colon X\to Y$ be a linear mapping. Denote its graph by $G(T)$ and let $p_1\colon X\times Y\to X$ and $p_2\colon X\times Y\to Y$ 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 $\{(x_i, Tx_i)\}$ in $G(T)$ which converges to $(x,y)\in X\times Y$ we have that $$x_i = p_1(x_i,Tx_i) \xrightarrow[i\to\infty]{} p_1(x,y) = x$$ and $$Tx_i = p_2(x_i,Tx_i) \xrightarrow[i\to\infty]{} p_2(x,y) = y,$$ by continuity of the projections. But then, since $T$ is continuous, $$Tx = \lim_{i\to\infty} Tx_i = y.$$ 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 $\tilde Tx = (x,Tx)$ 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.