= Solution
Let $(x_n)$ be a <Cauchy sequence> in the <graph norm>, and set $y_n=Tx_n$. Since
$$
\|y_n-y_m\|_Y\leq\|x_n-x_m\|_T,
$$
$(y_n)$ is a <Cauchy sequence> in the <Banach space> $Y$. Write $y_n\to y\in Y$.
Use the bounded <right inverse> to remove the part seen by $T$:
$$
w_n=x_n-Sy_n,\qquad Tw_n=Tx_n-TSy_n=y_n-y_n=0.
$$
Thus $w_n\in W$, and the <operator norm> bound gives
$$
\|w_n-w_m\|_X\leq\|x_n-x_m\|_X+\|S\|\|y_n-y_m\|_Y
\leq(1+\|S\|)\|x_n-x_m\|_T.
$$
The assumed <completeness> of $W$ supplies $w\in W$ with $w_n\to w$ in $\|\cdot\|_X$. Put $x=w+Sy\in X$. The <linearity> of $T$ and the <right inverse> identity give $Tx=y$. Therefore
$$
\|x_n-x\|_T\leq\|w_n-w\|_X+(1+\|S\|)\|y_n-y\|_Y\longrightarrow0.
$$
We have proved the <completeness of a graph norm from a bounded right inverse>:
$$
\boxed{(X,\|\cdot\|_T)\text{ is a Banach space}.}
$$
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