This is a question from the book: A Course in Multivariable Calculus and Analysis which I am self-studying. Also, it is my first question here and english is not my native language.
I must prove: If $D$ is compact, then $D$ is closed. (Hint: If $(x_0, y_0) \in D\backslash \partial D$, then the set of open squares centered at $(x, y)$ and of radius $|(x, y)−(x_0, y_0)|/2$, as $(x, y)$ varies over $D$, is an open cover of $D$.)
First of all I don't really know how this hint can help me. I can't see how the set of squares can contain the point $(x_0,y_0)$, because if $(x,y) = (x_0,y_0)$ then the radius is $0$, therefore the square centered at $(x_0,y_0)$ is the empty set. So I can't accept that it is an open cover of $D$.
Supposing it is an open cover of $D$, it must have a finite subcover by compactness of $D$. Which is impossible, because there's always a neighborhood of $(x_0,y_0)$ that isn't covered.
And finaly, what could I do with those $(x_0,y_0) \in D\backslash \partial D$?
Thank you for your answer.