You want the word "intersection" of sets, not "unity" (and NOT "union" which was my first guess). Other than that, your English is excellent.
That "union of any three is non-empty" is interesting. There are any number of examples of bad induction where the error is that you cannot go from n= 1 to n= 2 although it is easy to go from n to n+ 1 for n> 1.
And, of course, the whole bit about "convex subsets" is irrelevant. What is important is that if the intersection of any three sets in a collection of sets is non empty then the intersection of all of them is non-empty.
Let be a collection of n sets such that the intersection of any three of them is non-empty. Then n must be at least 3 for that to make sense.
Base case: n= 3. Since only contains 3 sets if "the intersection of any 3 is non-empty" then certainly the intersection of all 3 is non-empty.
Now suppose this is true for n= k. We will show it is true for n= k+1.
Suppose is a collection of k+1 sets having the property that the intersection of any 3 of the sets is non-empty.
Let U be any one of the sets and Let be with U removed. Then still has the property that the intersection of any three of its sets is non-empty and since we are assuming that the theorem is true for n= k, it is true that the intersection of all sets in is non-empty. Let X be that intersection.
Let V be any set other than U and let be with V removed. By the same argument, the intersection of all the remaining sets is non-empty. Let Y be that intersection.
Now V is a member of and U is a member of . And, since k is at least three, there exist a third set, W, which is in both and (this is the crucial point and why we need that "intersection of any three sets is non-empty) . All of the points in X are in both V and W and all of the points in Y are in both U and W. That means that W contains points of the intersection of X and Y is in W and so that interersection is non-empty. From that, it follows that the intersection of all sets in is non-empty.