Show that the collection of all open subsets of $X$ that are contained in $Y$ is a topology on $Y$. This question is from a text book. Please let me know if my proof is vaild. 

Suppose $X$ is a topological space and $Y$ is an open subset of $X$. Show that the collection of all open subsets of $X$ that are contained in $Y$ is a topology on $Y$.

If all open subsets of $X$ that are contained in $Y$ is not a topology on $Y$, then there is some open subset of $Y$ that is not contained in $X$. However, this contradicts the given "$Y$ is an open subset of $X$"
 A: Let $\mathcal Y$ be the set of all open subsets of $X$ that are subsets of $Y$.. You have to check the three conditions:


*

*$\varnothing \in \mathcal Y$, which is clear and because $Y$ is open in $X$, and it's a subset of itself, $Y \in \mathcal Y$.

*The Union of arbitrarily many elements of $\mathcal Y$ is again a subset of $Y$, and each being open in $X$, their union is open in $X$ as well, and so $\in \mathcal Y$.

*The intersection of ...
A: Let $\mathcal{F_Y}=\{O:O\in X,\:O\subset Y,\:Y\in X,\: \text{O and Y are open}\}$. 
Then for each $O\in\mathcal{F_Y}$, there is $O=O\cap Y$ for $O\subset Y$. We prove that $\mathcal{F_Y}$ is a topology in $Y$.
Let $O_{\alpha}$ be any collection of sets in $\mathcal{F_Y}$. Then $\bigcup O_{\alpha}$ is open for $O_{\alpha}$ is open and arbitrary union of open sets is open. Moreover
$$
\bigcup O_{\alpha}=\bigcup (O_{\alpha}\cap Y)=(\bigcup O_{\alpha})\cap Y\subset Y
$$
So $\bigcup O_{\alpha}\in\mathcal{F_Y}$.
Let $O_{i}$ be any finite collection of sets in $\mathcal{F_Y}$. Then $\bigcap_{i=1}^n O_i$ is open for $O_{i}$ is open and finite intersection of open sets is open. Moreover
$$
\bigcap_{i=1}^n O_i=\bigcap_{i=1}^n (O_i\cap Y)=(\bigcap_{i=1}^n O_i)\cap Y\subset  Y
$$
So $\bigcap_{i=1}^n O_i\in \mathcal{F_Y}$. 
Finally $\varnothing\in\mathcal{F_Y}$ for $\varnothing$ is open and $\varnothing\subset  Y$.
So $\mathcal{F_Y}$ is a topology in $Y$.
