In Green and Gamelins book "An Introduction to Topology" there is an exercise Id like some feedback on. If $Y$ is a subset of $X$, then prove that $\textbf{Int}(Y)$ is the union of all open subsets of $X$ that are contained in $Y$. I understand the books solution, which is, if we consider each $x \in U$ where $U$ is the union of open subsets in $Y$. There exists a ball $B(x, r) \subset U$, open. So each $x \in U$ is an interior point of $Y$ and $U \subseteq \textbf{Int}(Y)$. But $U$ is the set of all open subsets of $Y$, so $\textbf{Int}(Y) \subseteq U$ and thus $\textbf{Int}(Y) = U$.
Now, my solution goes as follows. Each open subset in $Y$ is a union of open balls. Around each point in $\textbf{Int}(Y)$ there exists an open ball, so $\textbf{Int}(Y)$ is open and each open subset of $Y$ is a subset of the union of these open balls. Taking the union of these subsets would exhaust the set of open balls in $\textbf{Int}(Y)$ and we are done.
My question is, is my solution enough?