= Solution
Both terms in the <graph norm> are finite and nonnegative. Moreover, $\|x\|_T=0$ forces $\|x\|_X=0$, hence $x=0$; conversely $T0=0$ gives $\|0\|_T=0$. This proves <definiteness of a norm>.
The <linearity> of $T$ and <absolute homogeneity of a norm> give, for each real $a$,
$$
\|ax\|_T=|a|\|x\|_X+\|aTx\|_Y=|a|\|x\|_T.
$$
Applying the <triangle inequality> in both <normed spaces> gives
$$
\|x+z\|_T=\|x+z\|_X+\|Tx+Tz\|_Y\leq\|x\|_T+\|z\|_T.
$$
Thus
$$
\boxed{\|\cdot\|_T\text{ is a norm on }X.}
$$
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