Note that you titled this "isolation points" but asked about "interior points". An "isolated point" of a set cannot be an interior point.
For example, if A= (0, 1), the set of all x such that 0< x< 1, the interior points are just points in A itself. That is true because:
if x in (0, 1) then 0< x< 1. Let d1= x, d2= 1- x. If d1< d2, the neighborhood (x-d1, x+d1) is a subset of A. If d2< d1, (x-d2, x+ d2) is in A.
If A= [0, 1], the set of all x such that 0\le x\le 1, the interior points are again the points in (0, 1). That's true because any neighborhood of "0", (-d, d), includes points outside A (negative numbers to -d) and any neighborhood of "1", (1-d, 1+ d), includes points outside A (numbers larger than 1 up to 1+ d) so "0" and "1", while in the set, are not interior[points].
Some other useful words: we say that point, p, is an "exterior" point of set A if and only if it is an interior point of the complement of A. The complement of (0, 1) is (-\infty, 0]cup [1, \infty) and the complement of [0, 1] is (-\infty, 0)\cup (1, \infty) both of which have (-\infty, 0)\cup(1, \infty) as interior points (so that the "exterior" points of both (0, 1) and [0, 1] are (-\infty, 0)\cup (1, \infty). The boundary points of a set are all points that are neither "interior points" nor "exterior points" of the set. Here, the boundary points of both (0, 1) and [0, 1] are the points "0" and "1".
The difference is that those boundary points are in [0, 1] and not in (0, 1). We say that (0, 1) containing none of its boundary points, is an "open" set and [0, 1], containing all of its boundary points, is a "closed" set.
Or course, a set may contain some of its boundary points but not all. (0, 1] is an example. Since neither "none of its boundary points" nor "all of its boundary points" is true, such a set is neither open nor closed.
Although it is unusual, it is possible for a set to have NO boundary points. In that case "none" and "all" are the same, such a set is both open and closed.