$A$ closed in $Y$ and $Y$ closed in $X$ , then $A$ is closed in $X$

Ok, I think I solved it:

If $Y-A$ is open, then it's equal to the intersection of an open set in $X$, let's call it $B$, with $Y$: $Y-A=Y\cap B$, and so $B$ contains $Y-A$, so $(X-Y)\cup B$ is open because it's the union of two open sets, and that is equal to $X-A$ because $B$ contains all $Y$ but $A$, and $X-Y$ contains all $X$ but $Y$.

Am I right?