More specifically, let $O_1, . . . , O_n$ be a finite collection of open subsets of the continuum, $C$. Then the intersection $O_1 ∩ · · · ∩ O_n$ is open as well. I think it is possible to do it without considering metric spaces (and therefore balls of radius $r$). Also, if you could prove why the intersect of an infinite amount of open subsets turns out to be possibly closed, it would be very much appreciated as this is difficult to wrap my head around.
Also open sets are defined as sets that don't contain their endpoints. Sorry for being unclear.
Best Answer
For your second question, for any positive integer $n$, let $A_n=(-1/n,1/n)$.
The intersection of all the $A_n$ is $\{0\}$.
Or else let $B_n=(0,1/n)$. The intersection of all the $B_n$ is the empty set, which is closed (and open).
The answer to the first question depends on the details of your definition of open. Let us define a set $A$ of reals to be open if for any $x\in A$, there is a positive $\epsilon$ (which usually depends on $x$) such that the interval $(x-\epsilon,x+\epsilon)$ is a subset of $A$.
Now let $x$ be in the intersection of your $O_i$. Then for every $i$, there is a positive $\epsilon_i$ such that $(x-\epsilon_i,x+\epsilon_i)$ is a subset of $O_i$. Let $\epsilon$ be the smallest of the $\epsilon_i$. Then the interval $(x-\epsilon,x+\epsilon)$ is a subset of $O_1\cap O_2\cap\cdots\cap O_n$.