Show that $E[ E[Z\lvert \mathcal{F}_{T}\lvert \mathcal{F}_{S}]=E[Z\lvert \mathcal{F}_{T\land S}]$

Let $Y$ be the random variable $Y=E[Z| \mathcal{F}_{T}]$. I am assuming we work with a complete filtration $(\mathcal F_t)$ on some space $(\Omega, \mathcal F,\Bbb P)$.

Then $(*)$ gives: $$ \begin{aligned} E[Z| \mathcal{F}_{T}] &=E[Z |\mathcal{F}_{T\land S}] \qquad\text{ on } A:=\{T\le S\}\ ,\\ E[Y| \mathcal{F}_{S}] &=E[Y |\mathcal{F}_{T\land S}] \qquad\text{ on } B:=\{S\le T\}\ ,\text{ so } \\[2mm] E[Z\cdot 1_A| \mathcal{F}_{T}] &=E[Z\cdot 1_A |\mathcal{F}_{T\land S}] \qquad(1) \\ E[Y\cdot 1_{\bar A}| \mathcal{F}_{S}] &=E[Y\cdot 1_{\bar A} |\mathcal{F}_{T\land S}] \qquad(2)\qquad\text{ so } \\[2mm] E[\ E[Z\cdot 1_A| \mathcal{F}_{T}]\ | \mathcal{F}_{S}] &= E[\ E[Z\cdot 1_A |\mathcal{F}_{T\land S}] \ |\mathcal F_S] \qquad\text{ by }(1)\\ &= E[Z\cdot 1_A |\mathcal{F}_{T\land S}]\\ &= E[Z |\mathcal{F}_{T\land S}]\cdot 1_A \\[2mm] E[\ E[Z\cdot 1_{\bar A}| \mathcal{F}_{T}]\ | \mathcal{F}_{S}] &=E[Y\cdot 1_{\bar A}| \mathcal{F}_{S}]\\ &=E[Y\cdot 1_{\bar A}| \mathcal{F}_{T\land S}] \qquad\text{ by }(2)\\ &=E[Y| \mathcal{F}_{T\land S}]\cdot 1_{\bar A}\\ &=E[\ E[Z| \mathcal{F}_{T}]\ | \mathcal{F}_{T\land S}]\cdot 1_{\bar A}\\ &=E[Z | \mathcal{F}_{T\land S}]\cdot 1_{\bar A}\\ \end{aligned} $$ And we add. (We have used $\bar A\subseteq B$.)

$\square$

What seems to be missing here is the following fact:

If $Y$ is $\mathcal{F}_{S}$-measurable, then $Y 1_{\{S < T\}\}}$ is $\mathcal{F}_{S \wedge T}$-measurable.

(You already seem to have noticed that $Y 1_{\{S \leq T\}}$ is $\mathcal{F}_{S \wedge T}$-measurable.)

Now notice that if $A \in \mathcal{F}_{S \wedge T}$, then $A \cap \{S < T\} \in \mathcal{F}_{S \wedge T}$ and, thus, \begin{align*} E[E[Z\mid\mathcal{F}_{S}] 1_{\{S < T\}} : A] &= E[Z : A \cap \{S < T\}] \\ &= E[E[Z \mid \mathcal{F}_{S \wedge T}] 1_{\{S < T\}} : A]. \end{align*} Thus, since $E[Z \mid \mathcal{F}_{S}]1_{\{S < T\}}$ is $\mathcal{F}_{S \wedge T}$-measurable by the observation above, the arbitrariness of $A$ implies that \begin{equation*} E[Z \mid \mathcal{F}_{S}] 1_{\{S < T\}} = E[Z \mid \mathcal{F}_{S \wedge T}] 1_{\{S < T\}}. \end{equation*}

Finally, notice that if $A \in \mathcal{F}_{S}$, then our observation gives $A \cap \{S < T\} \in \mathcal{F}_{S \wedge T}$ so we can write \begin{equation*} E[E[Z\mid\mathcal{F}_{T}] : A \cap \{S < T\}] = E[Z: A \cap \{S < T\}] = E[Z 1_{\{S < T\}} : A]. \end{equation*} By the arbitrariness of $A$ and the inclusion $\{S < T\} \in \mathcal{F}_{S}$, this implies \begin{equation*} E[E[Z\mid \mathcal{F}_{T}] \mid \mathcal{F}_{S}] 1_{\{S < T\}} = E[E[Z \mid \mathcal{F}_{T}] 1_{\{S < T\}} \mid \mathcal{F}_{S}] = E[Z \mid \mathcal{F}_{S}] 1_{\{S <T\}}. \end{equation*}

Combining the two identities obtained above, we find the equation we needed: \begin{equation*} E[E[Z\mid \mathcal{F}_{T}] 1_{\{S < T\}} \mid \mathcal{F}_{S}] = E[E[Z\mid \mathcal{F}_{T}] \mid \mathcal{F}_{S}] 1_{\{S < T\}} = E[Z \mid \mathcal{F}_{S \wedge T}] 1_{\{S <T\}}. \end{equation*}