Note: you need to assume (for instance) that the two connected subsets have a non-empty intersection.
Let be connected sets such that . Assume by contradiction that is disconnected. Then there would exist two disjoint non-empty open subsets of such that (nb: are open for the induced topology on ). Let , . Then are disjoint open subsets of such that . Similarly, we can define for . Since and are connected, we must have either or , and either or . Each case leads to a contradiction. For instance, if and , then , whereas . If and , then and , hence ( are disjoint), which contradicts since .
Because , , and are disjoint.
Because . You should draw a diagram if you don't get it. I just split into two parts according to the partition of . If cut in two disjoint pieces, then their intersections with cut it also in two disjoint pieces.Also, you said
Then U_1,U_2 are disjoint open subsets of X such that U_1\cup U_2=X.
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How do you know this fact that the union of U1 and U2 is X?