If $\tau_{X}$ is the topology in $X$ then the subspace topology in any $Y\subset X$ is the induced topology:
$$\tau_{Y}=\{T\cap Y:T\in \tau_{X}\}$$
This follows that the closed in $Y$ are:
$$\overline{\tau_Y}=\{Y\backslash (T\cap Y):T\in \tau_{X}\}$$
So if $Z\subset Y\subset X$ then $Cl_{Y}(Z)=Y\cap Cl_{X}(Z)$.
I wonder why it's worth the equality $Cl_{Y}(Z)=Y\cap Cl_{X}(Z)$?
Proposition. [2.2.1, Eng] Let $X$ be a topological space and $M$ a subspace of $X$. The set $A\subset M$ is closed in $M$ if and only if $A=M\cap F$, where $F$ is closed in $X$. The closure $\tilde{A}$ of a set $A\subset M$ in the subspace $M$ and the closure $\overline{A}$ of $A$ in the space $X$ are related by the equality $\tilde{A}=\overline{A}\cap M$.
Proof. If $A =M\cap F$, where $F =\overline{F}\subset X$, then $M\setminus A=M\cap (X \setminus F)$ and $A$ is closed in $M$ as the complement of an open set. Conversely, if $A$ is a closed subset of $M$, then $M\setminus A = M\cap U$, where $U$ is open in X. Thus $$A = M\setminus(M\setminus A) = M\setminus (M\cap U) = M\cap(X\setminus U),$$ and $A = M\cap F$, where $F = X \setminus U$ is closed in $X$.
By definition of the closure operator, $\tilde{A}$ is equal to the intersection of all closed subsets of $M$ that contain $A$, i.e., of all sets $M\cap F$, where $F = \overline{F}$ and $A\subset F$. This gives the equality $\tilde{A} = M\cap\overline{A}$. $\square$
References
[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.