Understanding a lemma regarding subspace topology I am learning topology and am struggling with some of the concepts of open sets regarding subspaces.
The problem I am working on says, that Y is a subspace of X, and A is a subset of Y, then show that the topology A inherits as a subspace of Y is the same topology it inherits as a subspace from X.
now what I am trying to prove is that A is a basis for the subspace topology of Y of X, and A is the basis for the subspace topology on X, then these two topologies are the same.
Here is the lemma I am struggling to utilize , it says, "Let Y be a subspace of X, if U is open in Y and Y is open in X, then U is open in X."
Since A is a subset of Y, does this imply that A is open in Y? (i think so).
Y is open in X, so does this imply that A is open in X? (I think so)
If A is open in Y, and A is open in X, does this mean that A is open in $Y \bigcap U$ for $U \subset X$ ?  (This I am not sure of).
Further, assuming my intuition is right,  if A is open in $Y \bigcap U$ does this mean for $x \in Y \bigcap U$ that $x \in A \bigcap U$ since A is open in both Y and X?  
this yields my needed argument saying that $A \bigcap U \subset Y \bigcap U$  this can then be shown that $\mathbb{B}_{A}$ is a basis for both topologies.
my question is that just because A is a subset of Y and A is open in X, and $x \in Y \bigcap U$ , I am not sure if this implies that $x \in A \bigcap U$.
Thank you again so very much and I will greatly appreciate any clarification.
 A: Answer on the problem mentioned in second alinea.
Let it be that $X$ is a set equipped with a topology $\tau$, and
let $A\subseteq Y\subseteq X$. 
Then $Y$ inherits a subtopology $\tau_{Y}=\left\{ Y\cap U\mid U\in\tau\right\} $
and likewise $A$ inherits a subtopology $\tau_{A}=\left\{ A\cap U\mid U\in\tau\right\} $.
Looking at $A$ as a subspace of $Y$ it also inherits a subtopology
$\tau'_{A}=\left\{ A\cap V\mid V\in\tau_{Y}\right\} $ and to be shown
is that $\tau_{A}$ and $\tau'_{A}$ coincide. 
If $O\in\tau'_{A}$
then $O=A\cap V$ for some $V\in\tau_{Y}$. From $V\in\tau_{Y}$ it
follows that $V=Y\cap U$ for some $U\in\tau$. Then $O=A\cap Y\cap U=A\cap U$
showing that $O\in\tau{}_{A}$. 
Proved is now that $\tau'_{A}\subseteq\tau_{A}$. 
Conversely let $O\in\tau_{A}$. Then $O=A\cap U$ for some $U\in\tau$.
Now note that we can also write $O=A\cap Y\cap U=A\cap W$ where $W:=Y\cap U\in\tau_{Y}$.
This shows that $O\in\tau'_{A}$. 
Proved is now that $\tau{}_{A}\subseteq\tau'_{A}$
and you are ready.
A: A basis is a collection of sets, so for a mere subset, you have to show that the subset is a collection of sets, not just a single set before showing that a subset can be a basis.
The second part of my confusion, assuming that A is open in Y and X, is that if one were to show that A were open in $Y \bigcap U$ where U $\subset X $ , one needs to show that A is contained in U.  both of these I do not have and can not assume.
thank you very much.
