Let $(S,\tau)$ be a topological space, and let $H\subseteq S$.
Using this definition, we can define the topological subspace $(H,τ_H)$ where $τ_H:=\{ U \cap H : U \in τ \}$.
Now, having just begun my independent study of topology, I found this definition to be a little surprising. Why not just define the subspace so that $τ_H:=\{U\in τ : U\subseteq H\}$?
On
Regarding your question (basically just saying the same as Pascal):
If $Y \not \in \tau$, then $Y \subset Y$ would not be open, which is an axiom for a topology. Therefore we have to define the subspace topology in a way which does not depend on properties of the subset.
The idea for the definition is the following:
If we have a topological space $(X,\tau)$ and a subset $Y \subset X$, then we want the inclusion map $i \colon Y \rightarrow X$ to be continuous. This means for every open set $U \subset X$ we need the preimage under the inclusion map $i^{-1}(U) = U \cap Y$ to be open in $Y$. We now take th smallest topology such that this is satisfied (which is just taking this as definition for open sets) and that is how the subspace topology is defined.
Recall that a topological space $(X,τ)$ must have the property that $X\inτ$. That is, $X$ itself must be "open."
The problem with defining $τ_H$ to be $\{U\in τ : U\subseteq H\}$ is that $(H,τ_Η)$ could only be a topology if $H$ is "open" in our original set $S$. We want a definition which allows us to take any (nonempty) subset of $S$ and form a topology.