Is this alternate answer right? This is really basic, but I don't want to get the fundamentals wrong.

Show that if $A$ is closed in $Y$ and $Y$ is closed in $X$, then $A$ is closed in $X$.

The solution uses subspaces (which I understand and agree),  but I did this instead.
$A$ is closed in $Y$, so $Y - A$ is open in $Y$. Also $X - Y$ is open in $X$, so $(X - Y)  \cup (Y - A) = A^c$ is open in $X$.

What you haven't shown is why $Y-A$ is open in $X$; since you want to claim that $A$ is closed in $X$.
Why is this true?
