= Solution
A <linear operator> has closed graph when $\{(x,Tx):x\in X\}$ is closed in $X\times Y$. The <closed graph theorem> states that an everywhere-defined <linear operator> between <Banach spaces> is bounded if and only if it has closed graph. Boundedness gives the closed-graph property by continuity. Conversely a closed graph is Banach in the product <norm> $\|(x,y)\|=\|x\|+\|y\|$. Projection of the graph onto $X$ is a bounded linear bijection, so the <open mapping theorem> makes its inverse bounded. Composing that inverse with the second-coordinate projection proves \b[$T$ is bounded].
For the composition assertion, suppose $x_n\to x$ and $Tx_n\to y$. Boundedness of $S$ gives $STx_n\to Sy$, and boundedness of $ST$ gives $STx_n\to STx$. Injectivity of $S$ forces $y=Tx$. Hence $T$ has closed graph and is bounded by the theorem.
For the derivative bound take $Y=C^1[0,1]$ with <norm> $\|v\|_{C^1}=\|v\|_\infty+\|v'\|_\infty$, and $Z=C[0,1]$ with its <supremum norm>; both are Banach. Regard $T(x)=Ux$ as a map into $Y$, and let $S:Y\to Z$ be inclusion. This inclusion is bounded and injective, and $ST=U$ is bounded by hypothesis. The preceding result gives a constant $c$ with $\|Ux\|_{C^1}\leq c\|x\|$. In particular \b[$|(Ux)'(t)|\leq c\|x\|$ uniformly in $t$].
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