Let's prove the first complication.
If $Y\subseteq X$ then we have to prove that $X\cup Y^c=\Omega$.
To prove it let's show both inclusions: it is clear that $X\cup Y^c\subseteq \Omega,$ so we only have to prove the other one. Take $x\in \Omega:$ if it belongs to $X$ we're done, otherwise it belongs to $X^c,$ that by hypothesis on $Y$ is a subset of $Y^c.$ So it certainly belongs to the union of $X$ and $Y^c.$
The other arrow is similar: if $X\cup Y^c=\Omega$ then we have to prove that $Y\subseteq X.$
By absurd, suppose that $\exists x\in Y\setminus X.$ Then it wouldn't belong to $X\cup Y^c$ because it is nor in $X$ neither in $Y^c.$ So it is an element of $\Omega$ which doesn't belong to $X\cup Y^c,$ which goes against the hypotheses.
The second statement is similar to prove, but if you want I'll make it.