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?

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 $Y.$
