Subbasis of a Subspace Let $X$ be a topological space, and let $\mathscr{S}$ be a subbasis for the topology of $X$; let $Y$ be a subspace of $X$, and let $\mathscr{S}_Y$ be given by
$$
\mathscr{S}_Y := \big\{ Y \cap S \, | \, S \in \mathscr{S} \big\}. 
$$
Then how to show that this collection $\mathscr{S}_Y$ is a subbasis for the subspace topology that $Y$ inherits as a subspace of $X$?
My Attempt:

The subbasis $\mathscr{S}$ is a collection of some subsets of $X$ such that
$$
\bigcup_{S \in \mathscr{S} } S = X. \tag{1} 
$$
Therefore we have
$$
\begin{align} 
\bigcup_{A \in \mathscr{S}_Y } A &= \bigcup_{S \in \mathscr{S} } (Y \cap S) \\ 
&= Y \cap \left( \bigcup_{S \in \mathscr{S} } S \right) \\ 
&= Y \cap X \qquad \mbox{ [ by (1) above ] }  \\ 
&= Y. 
\end{align}
$$
That is,
$$
\bigcup_{A \in \mathscr{S}_Y } A = Y. \tag{2} 
$$


Let $\mathscr{B}$ be the basis for the topology of $X$ determined by the subbasis $\mathscr{S}$. Then by definition
$$
\begin{align} 
& \qquad \mathscr{B} \\ 
&= \mbox{ the collection of all the finite intersections of sets in $\mathscr{S}$ } \\ 
&=  \left\{ \cap_{i=1}^n S_i \, | \, n \in \mathbb{N}, \, S_i \in \mathscr{S} \mbox{ for } i = 1, \ldots, n \right\}. 
\end{align} 
$$
And, the basis $\mathscr{B}_Y$ for the subspace topology on $Y$ determined by $\mathscr{B}$ is given by
$$
\mathscr{B}_Y = \big\{ Y \cap B \, | \, B \in \mathscr{B} \big\}.
$$
So we can write
$$
\begin{align} 
\mathscr{B}_Y &= \left\{ Y \cap \left( \cap_{i = 1}^n S_i \right) \, | \, n \in \mathbb{N}, \, S_i \in \mathscr{S} \mbox{ for } i = 1, \ldots, n \right\} \\
&= \left\{ \cap_{i=1}^n \left( Y \cap S_i \right) \, | \, n \in \mathbb{N}, \, S_i \in \mathscr{S} \mbox{ for } i = 1, \ldots, n \right\},
\end{align} 
$$
which shows that the basis $\mathscr{B}_Y$ consists of all the finite intersections of sets from the collection $\mathscr{S}_Y$, thus showing that the latter collection is indeed a subbasis for the subspace topology on $Y$.

Is this proof correct, clear, and sound enough in each and every detail? Or, are there any deficiencies therein?
 A: Your proof is okay but there is a more direct route (omitting bases) that makes use of a theorem that IMV should be known by everyone who learns topology.

Let me first state that in situations where $Z$ is some set and $\mathcal V$ is some collection of subsets of $Z$ I will use the notation $\tau(\mathcal V)$ for the smallest topology on $Z$ that contains $\mathcal V$ as a subcollection. Actually $\tau(\mathcal V)$ is the intersection of all topologies on $Z$ that contain $\mathcal V$. Further in this context $\mathcal V$ is a subbase of $\tau(\mathcal V)$.

Theorem:
If $X$ and $Y$ are sets, $f:Y\to X$ is a function and $\mathscr S$ is a collection of subsets of $X$ then:$$\tau\left(f^{-1}\left(\mathscr{S}\right)\right)=f^{-1}\left(\tau\left(\mathscr{S}\right)\right)\tag1$$

Now if we go for $Y\subseteq X$ and $f:Y\to X$ as inclusion then this tells us:$$\tau\left(\mathscr{S}_{Y}\right)=\left\{ Y\cap U\mid U\in\tau\left(\mathscr{S}\right)\right\}$$which is exactly the statement you are asked to proof.

The theorem is proved by means of two lemma's which are both straightforward to prove:

*

*Lemma1: if $\tau_{X}$ is a topology on $X$ then $f^{-1}\left(\tau_X\right):=\left\{ f^{-1}\left(V\right)\mid V\in\tau_X\right\} $
is a topology on $Y$.


*Lemma2: if $\tau_{Y}$ is a topology on $Y$ then $\left\{ V\in\mathcal{P}\left(X\right)\mid f^{-1}\left(V\right)\in\tau_{Y}\right\} $
is a topology on $X$.
The first lemma tells us that $f^{-1}\left(\tau\left(\mathscr{S}\right)\right)$
is a topology and from $f^{-1}\left(\mathscr{S}\right)\subseteq f^{-1}\left(\tau\left(\mathscr{S}\right)\right)$
it follows that: $$\tau\left(f^{-1}\left(\mathscr{S}\right)\right)\subseteq f^{-1}\left(\tau\left(\mathscr{S}\right)\right)$$
The second lemma tells us that $\left\{ V\in\mathcal{P}\left(X\right)\mid f^{-1}\left(V\right)\in\tau\left(f^{-1}\left(\mathscr{S}\right)\right)\right\} $
is a topology and from: $$\mathscr{S}\subseteq\left\{ V\in\mathcal{P}\left(X\right)\mid f^{-1}\left(V\right)\in\tau\left(f^{-1}\left(\mathscr{S}\right)\right)\right\}$$it
follows that: $$\tau\left(\mathscr{S}\right)\subseteq\left\{ V\in\mathcal{P}\left(X\right)\mid f^{-1}\left(V\right)\in\tau\left(f^{-1}\left(\mathscr{S}\right)\right)\right\}$$
or equivalently that: $$f^{-1}\left(\tau\left(\mathscr{S}\right)\right)\subseteq\tau\left(f^{-1}\left(\mathscr{S}\right)\right)$$
Proved is now that equality $(1)$ is valid.

Let me mention another handsome application of $(1)$:
If $X,Y$ are topological spaces, $f:Y\to X$ is some function and $\mathcal V$ is a subbase for the topology on $X$ then:$$f^{-1}(V)\text{ is open for every }V\in\mathcal V\implies f\text{ is continuous}$$
This because then: $$f^{-1}(\tau_X)=f^{-1}(\tau(\mathcal V))=\tau(f^{-1}(\mathcal V))$$ according to $(1)$ and: $$f^{-1}(\mathcal V)\subseteq\tau_Y\implies \tau(f^{-1}(\mathcal V))\subseteq\tau_Y$$
A: $(X, \tau) $ topological space.
$Y\subset X$
Claim :
$\mathscr{S}_Y := \big\{ Y \cap S \, | \, S \in \mathscr{S} \big\}$ form a subbasis of $(Y, \tau_Y)$
Choose any $U\in \tau_Y$
Then $U=Y\cap V$ for some $V\in\tau$
Since $V\in\tau $ and $\mathscr{S}$ is subbasis for $\tau$,
$V =\cup \cap_{j\in J } S_j$ ,where $[ S_j \in \mathscr{S} ] \text {and } J \text { is finite index set } $
Now, $U=Y\cap (\cup \cap_{j\in J } S_j) $
Hence, $U = \cup \cap_{j\in J}(Y\cap S_j) $
Hence, $U\in \tau_Y$ can be written as union of finite intersection of members of $\mathscr{S}_Y$
