O is open. So there exists an open ball B(x,epsilon) in O containing x, for some epsilon > 0. Pick such an epsilon. Now note that for some natural N and all n>N, |x_n - x|<epsilon and hence x_n is in B(x,epsilon) and also in O for all n>N.
Hence. . .
(obviously if we assume that x is not a limit point of O, that is it is in the interior of O, the assumption that O is open is unnecessary)