It is clear that the second condition implies the first, right? So, assuming the first holds, construct a sequence satisfying the second. Suppose (x_n) converges to p, and each x_n is not equal to p. So let y_1 = x_1, and since x_n converges to p, there is some x_k with k>1 and |x_k-p| < |x_1-p|/2. So let y_2 = x_k. Then just repeat the process to get a subsequence of distinct elements that converge to p.