Let $T = (X, Y, K, e, d)$ and $T' = (Y, Z, K', e', d')$ be cryptosystems, and define $T' \circ T = (X, Z, K \times K', e'', d'')$ where $e''_{(k, k')}(x)=e'_{k'}(e_k(x))$ and $d''$ is defined similarly.
The point is to prove or find a counterexample to the following statement: if $T' \circ T$ is perfectly secure, i.e. $\text{Pr}(X=x\mid Z=z)=\text{Pr}(X=x)$, then $T$ or $T'$ is perfectly secure.
I believe that the statement is true, but I haven't managed to prove it on my own. Here's my attempt:
$$\begin{align}\text{Pr}(X=x) &= \text{Pr}(X=x\mid Z=z) \\ \Leftrightarrow \text{Pr}(X=x) &= \frac{\text{Pr}(X=x, Z=z)}{\text{Pr}(Z=z)} \\ \Leftrightarrow \text{Pr}(X=x) &= \frac{\sum_{y\in Y}\text{Pr}(X=x, Y=y, Z=z)}{\text{Pr}(Z=z)} \\ \Leftrightarrow \text{Pr}(X=x) &= \frac{\sum_{y\in Y}\text{Pr}(X=x \mid Y=y)\;\text{Pr}(Y=y \mid Z=z)\;\text{Pr}(Z=z)}{\text{Pr}(Z=z)} \\ \Leftrightarrow \text{Pr}(X=x) &= \sum_{y\in Y}\text{Pr}(X=x \mid Y=y)\;\text{Pr}(Y=y \mid Z=z) \end{align}$$
It seems like this expression can only be true if $\text{Pr}(X=x \mid Y=y) = \text{Pr}(X=x)$ or if $\text{Pr}(Y=y \mid Z=z) = \text{Pr}(Y=y)$, which is what we're trying to prove here. But this isn't really enough, is it?
Perfect secrecy as stated is what we usually call as Shannon's secrecy. It is equivalent to The following categorization:
An encryption scheme satisfies perfect secrecy if for all messages $m_1, m_2$ in message space $M$ and all ciphertexts $c \in C$, we have $$\Pr_{k \leftarrow K}[Enc(K, m_1) =c] =\Pr_{k \leftarrow K}[Enc(K, m_2) =c].$$ where both probabilities are taken over the choice of $k \in K$ and over the coin tosses of the (possibly) probabilistic algorithm $Enc$.
From this, it can be shown that for perfect secrecy a cryptosystem must satisfy $|K| \geq |M|$.
Based on this, suppose $|X|=a, |K|=b$ and $|Y|=c, |K'|=d$, then for perfect secrecy of $T' \circ T$, we should have $bd \geq a$.
From this you want to check if this necessarily implies either $a \geq b$ or $c \geq d$. As you can see the conclusion is not guaranteed. Hence the statement is not true.