Problem. Let $\mathcal{M}$ be a topology on a set $X$ and $Y \subset X$. Then prove that $\mathcal{M}_Y = \{Y \cap O | O \in \mathcal{M}$} is defined to be the induced topology on $Y$.
I have to satisfy three axioms (open axioms). One of which is the union of open subsets. Formally, if $Q_i \in \mathcal{M}$ ($\mathcal{M}$ is a topology on $Y$) is open for all $i$ in some interval $I$ then $\bigcup_{i \in I} Q_i \in \mathcal{M}$.
I have proved this, but I think mine is a bit too hand-wavy. Any tips/help to make it more clear and or any glaring errors I didn't notice?
Proof: Let $Z \subset \mathcal{M}_Y$ and denote $\bigcup Z$ as the collection of all open sets such that $\bigcup Z \in \mathcal{M}_Y$. Set \begin{equation*} Z' = \left\{M \in \mathcal{M}~|~M \cap Y \subset \bigcup Z\right\}. \end{equation*} By definition of topology, $Z' \subset \mathcal{M}$. Now, let $O = \bigcup Z'$. Hence, by definition, $O \in \mathcal{M}$. Thus, for $M \in Z'$, \begin{equation*} O \cap Y = \left(\bigcup M \cap Y\right) \subset \bigcup Z. \end{equation*}
Now for all $N \in Z$ there exists a set $M \in \mathcal{M}$ such that $N = M \cap Y \subset \bigcup Z$. But then we also have the for all $N \in Z$ there exists a $M \in Z'$ such that $N = M \cap Y$. Then $M \subset O$, and consequently, $ N \subset O \cap Y$. Hence, it follows that $\bigcup Z \subset O \cap Y$. Therefore, $\bigcup Z = O \cap Y \in \mathcal{M}_{Y}$.