Why is the subspace topology defined as it is? Given a topological space $(X, \tau)$, and $Y \subset X$, the subspace topology on $Y$ is defined to be
$$\tau_Y = \{Y \cap U : U \in \tau\}.$$
The definition itself is clear and self-explanatory, what I am wondering here is why is it defined this why? Is it so that the collection of open sets within $Y$ do indeed form a topology?
 A: You can of course define many topologies $\tau_Y$ on the set $Y$, but a minimal requirement is certainly that the inclusion function $i : (Y,\tau_Y) \to (X,\tau_X)$ becomes continuous. Let us call such a topology admissible (this is just an ad hoc notation).
Obviously there exist a finest admissible topology $\tau_Y^f$ (the discrete topology) and a coarsest admissible topology $\tau_Y^c$ (take the intersection of all admissible topologies). Moreover, the set $\tau_Y^s = \{U \cap Y \mid U \in \tau_X\}$ occurring in your question is easily seen to be a topology on $Y$; it is admissible because $i^{-1}(U) = U \cap Y \in \tau_Y^s$ for all $U \in \tau_X$.
We have $\tau_Y^c = \tau_Y^s$. The inclusion $\tau_Y^c \subset \tau_Y^s$ holds by definition, and since $\tau_Y^c$ is admissible, we have $U \cap Y = i^{-1}(U) \in \tau_Y^c$ for all $U \in \tau_X$, thus $\tau_Y^s \subset \tau_Y^c$.
This explains the special role of $\tau_Y^s$ among admissible topologies: It is the coarsest admissible topology on $Y$.
Let us now change perspective. We call a topology $\tau_Y$ on $Y$ subspace topology if it has the following universal property:

A function $f : (Z,\tau_Z) \to (Y,\tau_Y)$, where $(Z,\tau_Z)$ is any topological space, is continuous if and only $i \circ f : (Z,\tau_Z) \to (X,\tau_X)$
is continuous.

That is, the continuity of functions into $(Y,\tau_Y)$ can be detected by regarding them as functions into the ambient $(X,\tau_X)$. This is a very reasonable approach to introduce a topology on $Y$.

*

*Each subspace topology  $\tau_Y$ is admissible:
The identity $id: (Y,\tau_Y) \to (Y,\tau_Y)$ is continuous, thus $i = i \circ id$ is continuous.


*There exists at most one subspace topology on $Y$:
If $\tau_Y, \tau'_Y$ are subspace topologies, then both $id : (Y,\tau_Y) \to (Y,\tau'_Y)$ and $id : (Y,\tau'_Y) \to (Y,\tau_Y)$ are continuous by 1. and the universal property. This means $\tau_Y = \tau'_Y$.


*$\tau_Y^s$ is a subspace topology (and thus by 2. the unique subspace topology):
The "only if" part is trivial because  $\tau_Y^s$ is admissible and for each admissible topology the continuity of $f$ implies the continuity of $i \circ f$. Let us prove the "if" part. Continuity of $i \circ f$ means that $(i \circ f)^{-1}(U) \in \tau_Z$ for all $U \in U_X$. But $(i \circ f)^{-1}(U) = f^{-1}(i^{-1}(U)) = f^{-1}(U \cap Y)$, thus $f^{-1}(V) \in \tau_Z$ for all $V \in \tau_Y^s$, i.e. $f$ is continuous.
Also have a look at an injective map is embedding iff its left composition with any continuous map is continuous.
If you know some category theory, Attempt to define the notion of subobjects will give you a broader understanding.
A: Some topics for you to think about
Metric spaces
Think of a metric space
\begin{equation*}
  d: X \times X \rightarrow X.
\end{equation*}

*

*What would be a good/natural metric to use in some $Y \subset X$?

*With this induced metric, what would be the neighbourhoods of a certain $a \in Y$?

*With this induced metric, what topology (open sets) does it give you?

*With this induced metric, when would it be true that $y_n \in Y$ converges to some $a \in Y$?

Restricting the co-domain
Given a continuous function $f: X \rightarrow Z$... if you restrict the co-domain to some $A \subset Z$ that contains $f(X)$...

*

*Should this new $\tilde{f}: X \rightarrow A$ be continuous?

*What a topology in $A$ would have to satisfy to make $\tilde{f}$ continuous?

Restricting the domain
Given a continuous function $f: X \rightarrow Z$... if you restrict the domain to some $A \subset X$...

*

*Should $f|_A$ be continuous?

*What would a topology in $A$ have to satisfy to make $f|_A$ continuous?

Thinking about nets
If you know about nets...

*

*A topological space is completely determined if you specify (with consistency) what nets converge and to where. Just like sequences in metric spaces.

*When exactly a net $y_\lambda \in Y \subset X$ should converge to some $a \in Y \subset X$?

*What topology in $Y$ gives (for $y_\lambda, a \in Y \subset X$)
\begin{equation*}y_\lambda \xrightarrow{\tau_Y} a \iff y_\lambda \xrightarrow{\tau_X} a?\end{equation*}
