Questions concerning Theorem 2.30 of Baby Rudin First question I have is the following:

Is [0, 1] open relative to  [0, 1]?

It seems open to me because for x in (0, 1), x is definitely an interior point of [0, 1] and for x=0, 1 there is a neighborhood centered at each point which is completely contained in [0, 1] because x>1 or x<0 are not in our attention.
If the answer is positive to the previous question, then I have this further question.
Theorem 2.30 of Baby Rudin is stated as follows.

Suppose $Y\subset X $. A subset E of Y is open relative to Y if and only if $E=Y \cap G $ for some open subset G of X.

If we consider a specific case like Y=[0, 1], X=$\mathbb {R} $ and E=Y, then G can be Y, so should G always be an open set?
My last question is the following. Is the purpose of implementing the new set G in the above Theorem to 'delete' all the points in X which are not in the sets containing the set E? Thus when G is intersected with Y, the result is the set E?
 A: The answer to your first question is yes.
As for the second question, the definition is correct as stated. Just because $Y$ is the intersection of $Y$ and a non-open subset of $X$ ($Y$ in this case) does not mean there isn't an open subset $G$ of $X$ such that $Y$ is the intersection of $G$ and $Y$ (any open superset of $Y$ will do here.)
For the third question, I'm not sure I understand exactly, but it seems this is at least on the right track. We are restricting to $Y$ and part of the idea is to guarantee $Y$ is open in $Y.$
A: *

*Yes. The whole space is always a member of a topology (that is, an open set). So $[0,1]$ is open in any topology on $[0,1]$. 

*G needs to be open in the topology on X. 

*You can say that the idea is indeed to “delete” (disregard) points not in Y. 

A: It might be helpful to keep the definition of a topological space in mind here. Remember that a topological space is a set $X$, together with a collection of subsets which satisfy some properties. 
To say that a set $Y$ is open relative to another set $X$ is to say that $Y$ is among the collection of subsets which form a topology on $X$. 
So your first question, "is $[0,1]$ open relative to $[0,1]$?" The answer is yes because (as part of the axioms of a topology) the whole space $[0,1]$ must be an open subset of $[0,1]$ (this is true for any topological space).
For your second question, what is the theorem saying? It states that for a fixed subset $Y\subset X$. A subset $E\subset Y$ is open (in the space $Y$!) if there is a $G\subset X$ open (in the space $X$!) such that $E=G\cap Y$. So in your specific example, $Y=[0,1]$, $X=\mathbb{R}$ and $E=[0,1]$, you cannot choose $G=[0,1]$, because that is not an open subset of $\mathbb{R}$. However, it is open in $Y=[0,1]$ because
$G=\mathbb{R}$ is open in $\mathbb{R}$ and $G\cap Y=[0,1]$. 
