Suppose that

is a separable metric space. Then there exists a sequence

such that it is dense in X. Prove that any open set

is a union of elements from the set

.

My idea so far.

So since X is separable, I know that every nonempty open subset U contains at least one element of the sequence

.

So say that

is an nonempty open subset in X, then there should exist at least one element from the sequence, say

, that is contained in U. And since U is open, we can find some

such that

.

Also, I believe that I can pick a point, say

, such that

for some small

. Then I should be able to find another point

since it is open. Then I can get another

...

But how should I proceed to get

?