Yes, you are right there's a problem, you need to show that for every y,z in Int(S) the line segment joining them is also in Int(S).

There's a famous proof that does it, you first pick a disk around y, if z is contained in this disk then you are done, if not then the point ay+(1-a)z is in this disk (for some a in [0,1]), not there's another disk with centre around this point, if z is contained in this disk then we are done if not procceed as in the preceding step, this problem should halt in a finite number of steps with the last point being z and so the line joining y and z is in the interior of S as well.

For the limit points, you can use the fact that there is a sequence of points in E which converge to the limit point, for example suppose:2. Prove that the closure of a convex set is convex.

My attempt, What I have so far: Let be the closure of . Pick arbitrary , I need to show .

3 cases:

(i) ; I showed the implication from here very easily.

(ii) , where is the set of all the limit points of .

(iii) Without loss of generality, and .

I'm having problem with the last "2" cases as I don't know how to proceed from the fact that something is a limit point to showing that .

Any suggestions would be appreciated. Thanks, in advance.

x is a limit point and y is a limit point, then for each of them, there are sequence points converging to them: {x_n}, {y_n}, so the second problem is translated to the third problem of proving that if one is a limit point and the other is interior point, in this case just look at

where y_n are interior points then for each n these line segment are in E, now if the limit of this sequence of lines was crossing cl(E) with its exterior then at least one of the other lines in the sequence of lines will cross cl(E) with its exterior, thus not contained as a whole in cl(E) opposite to our assumption.