I found a proof to the question, but mine is completely different (sort of). Is this correct?

If $X$ were Hausdorff, then consider $u,v \in X$ with disjoint neighbourhoods $U, V$ that separates the points. But this would mean $(U \cap V)^c = \emptyset^c \iff U^c \cup V^c = X$, implying that the union of two finite sets equals $X$, an infinite set. So $X$ cannot possibly be Hausdorff.

The proof I read strategies by showing one of the disjoint open set cannot be in the complement of the other, which is a finite set. But this contradicts the other one being in the topology.

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    $\begingroup$ Your proof is perfectly correct, except for misspelling "Hausdorff" twice. $\endgroup$ – bof Aug 23 '14 at 0:07
  • $\begingroup$ Not twice.... Thrice.... As you yourself corrected three times. $\endgroup$ – Pranita Gupta Feb 1 at 3:53

This proof is certainly correct.

  • $\begingroup$ Since I appear to be unable to comment under the main post, may I ask what exactly you were unsure about in regards to the correctness? $\endgroup$ – Michael Aug 23 '14 at 0:51

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