If $H$ is an open set in $Y$, there must be an open set $U$ in $X$ such that $H=U\cap Y$. We want to show that $A\cap\operatorname{cl}_X(Y\setminus H)=\varnothing$.
Now $Y\setminus H=Y\setminus(U\cap Y)=Y\setminus U$. (You may have to think a little about that last step. $Y\setminus(U\cap Y)=\{y\in Y:y\notin U\cap Y\}$, and $Y\setminus U=\{y\in Y:y\notin U\}$; do you see why those must be equal?)
$Y$ is closed in $X$, and $U$ is open in $X$, so $Y\setminus U$ is closed in $X$. (Why?) Thus, $Y\setminus H$ is closed in $X$, and therefore $\operatorname{cl}_X(Y\setminus H)=Y\setminus H$. Thus, $A\cap\operatorname{cl}_X(Y\setminus H)=A\cap(Y\setminus H)$, and we want to show that this is empty. But $H\supseteq A\cap Y$, so ... ?
Here’s a diagram that you may find helpful: vertical shading is $H$, horizontal shading is $A\cap Y$. $Y\setminus H=Y\setminus U$ is the region between the outline of $Y$ and the shaded set $H$.
