Does this theorem for bases also hold for subbases? Assume that we have a toological space $X$ with toplogy $\mathcal{T}$. If Y is a subspace of X, then $\mathcal{T}_Y=\{Y\cap U|U \in  \mathcal{T}\}$ is a topology on Y (that it really is a topology, has to be proved, I won't do that).
There is a theorem that says:

If $\mathcal{B}$ is a basis for the topology $\mathcal{T}$, then $\{B\cap Y|B \in \mathcal{B} \}$ is a basis for
  $\mathcal{T}_y$.

I am wondering if the same results hold for subbases? That is:

If $\mathcal{B}$ is a subbasis for the topology $\mathcal{T}$, then $\{B\cap Y|B \in \mathcal{B} \}$ is a subbasis for
  $\mathcal{T}_y$?

I tried proving this, but I am not sure if it is correct, or how to finish?
First we have that $\{B\cap Y|B \in \mathcal{B} \}$ really is a subbasis, all we have to do to show this is that the union is Y, but since the union of $\mathcal{B}$ is X since it is a subbasis on X, we have that the union of $\{B\cap Y|B \in \mathcal{B} \}$ is Y, so we atleast have a subbasis on Y.
So we need to check that the subbasis really generates our topology. Let $\mathcal{T}_Y'$ be the topology that the subbasis generates. Every element in $\mathcal{T}_Y'$ is a  union of finite intersections of elements in  $\{B\cap Y|B \in \mathcal{B} \}$, since each B is in $\mathcal{T}$ every $B\cap Y$ is in $\mathcal{T}_Y$ by definition, since $\mathcal{T}_Y$ is a topology, it is closed under unions of finite intersections. Hence $\mathcal{T}_Y'\subset\mathcal{T}_Y$
So what needs to be proved is that $\mathcal{T}_Y\subset\mathcal{T}_Y'$, and here I get some trouble. Let $Y\cap U\in \mathcal{T}_Y$. Since U is in $\mathcal{T}$ it is a union of finite interesection of elements in $\mathcal{B}$. But what I then need is that:
$U$ is a union of finite intersections of elements in $\mathcal{B}$, therefore $U\cap Y$ is the union of finite intersections of elements in $\{B\cap Y|B \in \mathcal{B} \}$. But how do I prove this? Or is it maybe not true?
 A: You could use distributivity of $\cap$ over $\cup$. This holds even for arbitrary unions.
Edit:
From the fact that $U$ is an arbitrary union (indexed by some set $I$) of finite intersections of elements from $\mathcal{B}$, it follows that:
$$U \cap Y = Y \cap \Big( \bigcup_{i\in I} \big( \bigcap_{0\leq j \leq n_i} O_{i,j} \big) \Big) = \bigcup_{i\in I}\Big( Y \cap \big( \bigcap_{0\leq j \leq n_i} O_{i,j} \big ) \Big) = \bigcup_{i\in I}\Big( \bigcap_{0\leq j \leq n_i} \big( Y \cap  O_{i,j} \big ) \Big)$$
A: It is true; the difficulties in proving it are more notational than conceptual, and if you like, I can show you. However, there’s an easier way to tackle the problem: use the known result for bases. This lets you deal only with finite intersections: the arbitrary unions have been dealt with in proving the theorem about bases.
You have a space $X$, a subbase $\mathscr{S}$ for the topology $\mathscr{T}$ on $X$, and a subspace $Y$ with subspace topology $\mathscr{T}_Y$, and you want to show that $\mathscr{S}_Y=\{S\cap Y:S\in\mathscr{S}\}$ is a subbase for $\mathscr{T}_Y$. Let $\mathscr{B}$ be the base for $\mathscr{T}$ generated by $\mathscr{S}$ (i.e., the set of all finite intersections of elements of $\mathscr{S}$). Let $\mathscr{B}_Y=\{B\cap Y:B\in\mathscr{B}\}$; you already know that $\mathscr{B}_Y$ is a base for $\mathscr{T}_Y$, so you’re done if you can show that $\mathscr{B}_Y$ is the base generated by $\mathscr{S}_Y$, i.e., that $B\in\mathscr{B}_Y$ if and only if there is a finite $\mathscr{F}\subseteq\mathscr{S}_Y$ such that $B=\bigcap\mathscr{F}$.
Suppose first that $\mathscr{F}$ is a finite subset of $\mathscr{S}_Y$, and let $B=\bigcap\mathscr{F}$. For each $S\in\mathscr{F}$ there is an $S'\in\mathscr{S}$ such that $S=S'\cap Y$, so
$$B=\bigcap\mathscr{F}=\bigcap_{S\in\mathscr{F}}(S'\cap Y)=\left(\bigcap_{S\in\mathscr{F}}S'\right)\cap Y\in\mathscr{B}_Y\;,$$
since $\bigcap_{S\in\mathscr{F}}S'\in\mathscr{B}$.
Now suppose that $B\in\mathscr{B}_Y$. Then there is a $B'\in\mathscr{B}$ such that $B=B'\cap Y$. By the definition of $\mathscr{B}$ there is a finite $\mathscr{F}'\subseteq\mathscr{S}$ such that $B'=\bigcap\mathscr{F}'$. Use this to find a finite $\mathscr{F}\subseteq\mathscr{S}_Y$ such that $B=\bigcap\mathscr{F}$, thereby completing the proof.
