Please go to this lecture(1). I have numbered my questions for organization and added the picture which I will constantly refer to. I also apologize in advance for this lengthy question
Right about here, he says that the point (the top one "enclosed" by the semi-circle) isn't open in $X$. I don't understand this. Why exactly isn't that open in $X$
Also he later talks about the semi-circle neighborhood being in $Y$ and that makes the point on the boundary an interior point. This confuses me. He left out the other part of the circle and just ignored that the part of the real "ball" lies outside $Y$ and therefore that ball isn't contained in $Y$.
Also are the sets $X$ and $Y$ even open sets in the lecture? Because later in the follow up video he writes $E = Y \cap G$. How can a metric space (is it even a set?) intersect with a set? Aren't they two different things? Isn't like saying $5 \cap [0,1]$? Sorry if the question is ignorant here.
Later in here, he tries to prove that $E = Y \cap G \iff E$ is open. He says that if a neighbourhood is contained in G, then the intersection of that neighbourhood and $Y$ is a neighbourhood in both $Y$ and $E$. But what happens if he takes a point that's not even in $Y$? And draw a neighbourhood around it such that it doesn't even intersect $Y$ or $E$? And suppose $Y$ really is a set (assuming (3) had been answered and clarifed), How does that imply this mysterious set intersection with an open set gives me and open set? That can't be true in general. What if $(-2,2)\cap [-1,1] = [-1,1]$? That doesn't give me an open set
Finally, I don't understand the idea and strategy in the forward direction of the proof. Like at all...
I found this proof (part of (5)) , in one line they say that $d(p,q) < r_p, q \in Y \implies q \in E$. How does that make sense? What if $q$ is outside of $E$?