$\mathbb Q$ is totally disconnected. What is the open set in subspace of $\mathbb Q$? I am trying to understand the proof that $\mathbb Q$  is totally disconnected.
If $Y$ is a subspace of $\mathbb Q$ containing two points, $p$ and $q$, we can choose irrational a lying between $p$ and $q$ such that we can write $Y$ as the union of the open sets in $Y$.
What I do not understand is the open sets in $Y$.
Since $Y$ is subspace of $\mathbb Q$ , any open set in $Y$ is intersection of $Y$ with open sets in $\mathbb Q$ . In fact, I think $\mathbb Q$  is not even an open set? 
How can we construct two open sets in Y that covers entire rational points?
 A: We have $Y \subseteq \mathbb{Q} \subseteq \mathbb{R}$. 
The topology on $\mathbb{Q}$ is the subspace topology from $\mathbb{R}$, which means that $O \subseteq \mathbb{Q}$ is open, iff there is an open subset $O' \subseteq \mathbb{R}$ such that $O' \cap \mathbb{Q} = O$.
And $Y$ has the subspace topology again with respect to $\mathbb{Q}$, so $U \subseteq Y$ is open iff there is an open subset $U' \subseteq \mathbb{Q}$ such that $U' \cap Y = U$.
It turns out that we can just directly say that $Y$ has the subspace topology with respect to $\mathbb{R}$ as well (so being a subspace of, in a topological sense, is also transitive): If we have $U$ and $U'$ as before, then for $O = U'$ we can find $U'' \subseteq \mathbb{R}$ with $U'' \cap \mathbb{Q} = U'$ and then 
$$U'' \cap Y = U'' \cap (Y \cap \mathbb{Q}) = (U'' \cap \mathbb{Q}) \cap Y = U' \cap Y = U  $$
and so $U$ is open in the subspace topology of $Y$ with respect to $\mathbb{R}$.
And if $O$ is open in $\mathbb{R}$ then $O \cap \mathbb{Q}$ is open in $\mathbb{Q}$, and then $(O \cap \mathbb{Q}) \cap Y$ is open in the subspace topology of $Y$ with respect to $\mathbb{Q}$. But this set equals $O \cap Y$ again (as $Y \cap \mathbb{Q} = Y)$ and so $O \cap Y$ is open in $Y$, both directly from being a subspace of $\mathbb{R}$ and as a subspace of $\mathbb{Q}$.
So in short, if we have $Y \subseteq \mathbb{Q} \subseteq \mathbb{R}$, then $Y$ gets exactly the same topology as a subset of either $\mathbb{Q}$ or $\mathbb{R}$ (when $\mathbb{Q}$ itself has the subspace topology with respect to $\mathbb{R}$, which is the case).
So if $a,b \in Y$ and $p$ is irrational between $a$ and $b$ then we can use $O_1 = (\leftarrow, p) \cap Y$ and $O_2 = (p,\rightarrow) \cap Y$, which are subspace open in $Y$ and cover them (as $p$ is the only possibly missed point but $p \notin \mathbb{Q}$ so in particular, $p \notin Y$). Both are non-empty as one contains $a$ and the other $b$.
