I think the problem is due to that, the randomized process you define is dependent of $X$. I think the following slightly modified version of your statement is correct:
Consider any two random variables $X$ and $Y$ distributed over some finite universal set $U$. Consider a randomized process $f(t, z)$, which is a (deterministic) function
$$ f: U\times R\to U, $$
where $R$ is the set of all random seeds.
Let $Z$ be a random variable distributed over $R$ that is independent of $X$ and $Y$.
Then
$$\Delta(f(X, Z); f(Y, Z))\leq \Delta(X; Y).$$
Note that letting the range of $f$ to be $U$ is without loss of generality. In fact, we can also let $R=U$ w.l.o.g..
Here is my proof:
For all $u\in U$ , let set $S_u\subseteq U\times R$ be $$ S_u\overset{\text{def}}{=}\{(t, z)\in U\times R: f(t, z) = u\}. $$
Let set $S_{u|z}$ be $$ S_{u|z} \overset{\text{def}}{=}\{t\in U:f(t, z)=u\}. $$
Note that for different $u_1\neq u_2$ , $S_{u_1|z}$ and $S_{u_2|z}$ are
disjoint. Hence the disjoint union of all $S_{u|z}$ is a subset of
$U$ , i.e. $$ \bigcup_{u\in U} S_{u|z}\subseteq U. $$
What we want
to prove is
$$ \begin{align}2\Delta(f(X, Z); f(Y, Z))&=\sum_{u\in U}\left|\Pr[f(X, Z)=u]-\Pr[f(Y, Z)=u]\right|\\
&=\sum_{u\in U} \left|\Pr[(X, Z)\in
S_u]-\Pr[(Y, Z)\in S_u]\right|\\ &=\sum_{u\in U}\left|\sum_{z\in
R}\Pr[Z=z]\cdot \left(\Pr[X\in S_{u|z}]-\Pr[Y\in
S_{u|z}]\right)\right|\\ &\leq\sum_{u\in U}\sum_{z\in
R}\Pr[Z=z]\cdot \left|\Pr[X\in S_{u|z}]-\Pr[Y\in S_{u|z}]\right|\\
&=\sum_{z\in R}\Pr[Z=z]\cdot\left(\sum_{u\in U} \left|\Pr[X\in
S_{u|z}]-\Pr[Y\in S_{u|z}]\right|\right)\\ &\leq\sum_{z\in
R}\Pr[Z=z]\cdot\left(\sum_{t\in U} \left|\Pr[X=t]-\Pr[Y=t]\right|
\right)&(*)\\
&=2\Delta(X;
Y). \end{align} $$
In step $(*)$, we take the union of all (disjoint) $S_{u|z}$, which is contained in $U$.
This finishes our proof.
Hope this helps!
