Let $X$, $Y$, and $Z$ be normed spaces, either all real or all complex. Let $T \colon X \longrightarrow Y$ and $S \colon Y \longrightarrow Z$ be bounded linear operators. Let $X^\prime$, $Y^\prime$, and $Z^\prime$ denote the dual spaces of (i.e. the normed spaces of all the bounded linear functionals defined on ) $X$, $Y$, and $Z$, respectively. Let $T^\times \colon Y^\prime \longrightarrow X^\prime$ and $S^\times \colon Z^\prime \longrightarrow Y^\prime$ denote the adjoint operators of $T$ and $S$, respectively. By definition, the operator $T^\times \colon Y^\prime \longrightarrow X^\prime$ is defined by $$ \left( \, T^\times (g) \, \right) (x) = g\big( T (x) \big) \ \mbox{ for all } \ x \in X$$ for all $g \in Y^\prime$.
This $T^\times$ is a bounded linear operator with $$\left\lVert T^\times \right\rVert = \lVert T \rVert.$$
How to show that $$\big( S \circ T \big)^\times = T^\times \circ S^\times?$$
My Attempt:
Let $h \in Z^\prime$. We need to show that $$ \big( S \circ T \big)^\times (h) = \left( T^\times \circ S^\times \right) (h).\tag{1} $$
Since $S \circ T \colon X \longrightarrow Z$, therefore $\big( S \circ T \big)^\times \colon Z^\prime \longrightarrow X^\prime$.
Thus, in order to show that the (1) above holds, we must show that $$ \left( \, \big( S \circ T \big)^\times (h) \, \right) (x) = \left( \, \left( T^\times \circ S^\times \right) (h) \, \right) (x) \ \mbox{ for all } \ x \in X. \tag{2} $$
But we note that $$ \begin{align} \left( \, \big( S \circ T \big)^\times (h) \, \right) (x) &= h \big( \, \big( S \circ T \big)(x) \, \big) \\ &= h\big( \, S \big( T(x) \big) \, \big) \\ &= \left( \, S^\times (h) \, \right)\big( T(x) \big) \\ &= T^\times \left( \, S^\times (h) \, \right)(x) \\ &= \left( \, \left( T^\times \circ S^\times \right)(h) \, \right)(x). \end{align} $$
Is this reasoning correct? If not, then where does the problem lie?