= Solution
The two recursions use a <synchronous coupling>: they receive exactly the same fresh inputs. Subtraction cancels the additive term:
$$
Z_n-X_n=U_n(Z_{n-1}-X_{n-1}).
$$
The variable $U_n$ is independent of the preceding difference and has <expected value> $1/2$. Thus <expectation of a product of independent random variables> gives
$$
\mathbb E|Z_n-X_n|=\frac12\mathbb E|Z_{n-1}-X_{n-1}|
=2^{-n}\mathbb E|c-E_0|\leq2^{-n}(|c|+1).
$$
By <Markov's inequality>, for every $\varepsilon>0$,
$$
\boxed{\mathbb P(|Z_n-X_n|>\varepsilon)\leq\frac{|c|+1}{\varepsilon\,2^n}\longrightarrow0.}
$$
This proves the requested <convergence in probability>; in fact the difference has <L1 convergence>. It is <synchronous contraction of a random affine recursion>, and the estimate holds for every real $c$.
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