1) Prove the union of a family of open intervals which intersection is not empty is an open interval.

2) Prove every open set of real numbers can be expressed (in just one way) as a numerable union of open intervals disjoint for pairs.

3) If an open interval is the union of two disjoint open sets then one of them is empty and the other is the whole interval.

4) Let A, B be subsets of real numbers then:

1. $\mathrm{int}(A \cap B) = \mathrm{int} (A) \cap \mathrm{int}( B)$
2. $\mathrm{int} (A) \cup \mathrm{int}(B) \subset \mathrm{int}(A \cup B) $

5) Let $A$ be a subset of $X$. And let $\partial A$ be the boundary of $A$. Prove:

$$X = \mathrm{int}(A) \cup \mathrm{int}(X \setminus A) \cup \partial A$$

So, for the first i tried a direct proof, i took two diferent points(a and b) of the union and then try to proof all the points between them are in the union too. 
We know a and b are in one of the open intervals, let A and B be those intervals, so a belongs A and b belongs B. Suppose a < b . Let c be any point such that a < c < b. I must prove c belongs to the union. Then i think maybe i could use the hypotesis about the no empty intersection of the family but i dont know how. I am stucked.

For the second the hint is use the first exercise. 

For thethird i tried contradiction so i supossed no one is empty and no one is the whole open interval. Since the union of the two open sets is the whole open interval then neither of the two open sets can be bigger than the interval, so the two open sets are two open disjoint no empty sets smaller than the interval. If we call them A and B we can suppsed that every element of A is smaller tha every element of B so the infimum(we can call it R) of B must be the supremum of A ad then must be bigger than any element of A. So R dont belongs to A and dont belongs to B. But belongs to the interval and this is the contradction we are lookng for. But i dont know if this is right and how to write it formally.
For 4 and 5 i am clueless.