circle cannot be realised as the union of two open sets with connected intersection. I wanted to find the fundamental group of circle using Van-Kampen theorem. 
I tried a lot to find two open sets $U$ and $V$ such that $U\cap V$ is path connected. Later on I read on Wikipeida that it is not possible  realize circle  as the union of two open sets with connected intersection. 
I want to prove it rigorously.
Is there a generalization of Van-Kampen theorem to find the fundamental group of circle ?
Is there way to resolve the issue ?
 A: Assuming $\pi_1(S^1)$ is nontrivial, here is a reason why the van Kampen theorem for fundamental groups cannot apply. Suppose $U,V$ are connected subsets of $S^1$ which form a open cover with connected intersection, with neither $U$ nor $V$ being all of $S^1$.  Using facts about open subsets of $\mathbb{R}$, it follows that $U$ and $V$ are both homeomorphic to an interval.  The van Kampen theorem would say that $\pi_1(S^1)$ is the amalgamated product of $\pi_1(U)$ and $\pi_1(V)$, but both of these groups are trivial, hence $\pi_1(S^1)$ would be trivial.
General idea: The van Kampen theorem applies to the case when all the generators come from the individual pieces, but not in the way the pieces are stitched together.  The between space is only allowed to add extra relations.
A relevant example for why open sets are needed: consider $U$ and $V$ being an open and half-open arc, respectively, so that $U\cup V=S^1$ and $U\cap V$ is connected.  These satisfy the hypotheses of the van Kampen theorem except for openness, yet with them we would conclude $\pi_1(S^1)$ is trivial.
A: You want the concept of the fundamental groupoid, for which one can handle non-connected intersections. To prove that the circle cannot be realised as the union of two open, proper subsets whose intersection is connected, note that the only connected open subsets of the circle are connected open arcs, hence at least one of the open sets must be the entire circle, hence not a proper subset.
