It is equivalent to make the same statement about the base open sets containing p, or the base open sets assigned to p, assuming the topology has a base. After all, every open set contains one of these base sets.
Clearly a limit point of the sequence s is a cluster point of s. And a cluster point of s is a limit point of the set s.
Let t be a subsequence of s. If p is a limit point or a cluster point of t then p is a cluster point of s.
In contrast, consider the line with two origins. (This space is not hausdorff.) Both p and q are limits of the reciprocal sequence sn = 1/n.
If the set R takes the form of a sequence, this transformation need not produce a subsequence. Let R3 = p and let all the other points in s be far away from p. yes, p is a limit of the set R, and the new sequence s has p in every position, yet s is not a subsequence of R. However, if p is a cluster point of the sequence R there is a convergent subsequence. Reconstruct s as above. When O is the intersection of the first n base open sets, there are infinitely many terms of R to choose from. Select one that is beyond the last one chosen, thus building a convergent subsequence.
In summary, the following holds in a first countable space.
Start with the positive integers and let a set be open if the number of integers in the set below n divided by n approaches 1. It is easy to show the union of open sets is open. Finite intersection is a little tricky, but not too bad. If n is large enough so that two sets include 99% of the numberrs below n, their intersection contains at least 98% of the numbers below n. Apply this reasoning for 1-ε and indeed the finite intersection is another open set.
Add 0 to the space and to each nonempty open set. The topology is still valid.
Let s be the sequence of positive integers. That is, sn = n. The point 0 is in every open set that intersects s; in fact it is in every nonempty open set. Thus 0 is a limit of the set s.
View s as a sequence and let O be an open set containing 0. Clearly O is not finite, hence it contains infinitely many points of s. Thus 0 is a cluster point of s.
Suppose there is a convergent subsequence t taken from s. We will build an open set that excludes infinitely many terms of t. If t has finitely many distinct integers, let an open set be the complement of t. This excludes all of t. Otherwise throw in the integers 1 through 100, and then all subsequent integers until you reach something in t. Leave this integer out. Then throw in the next thousand integers, and all subsequent integers, until you reach another term in t, which we will leave out. Do this for ten thousand, 100 thousand, a million, and so on. The result is an open set that omits infinitely many terms of t. Thus 0 is not the limit of t. similarly, a general point p cannot be the limit of t. Add p to the constructed open set and it is still open, and still misses t infinitely often. Thus the subsequence t has no limit.
The above is a round about way of proving the topology is not first countable. Let's prove it more directly, simply to confirm the fact. Suppose 0 has a countable base. Take the first open set in the base, find n so that 91% of the elements below n are present, make n at least 100 if it isn't already, and fold these first n elements into a set s; but leave one of them out. Now s is at least 90% full. Move to the second open set and advance n to 100 times its previous value, or more, so that the second open set is 99.2% full. Fold the elements beyond the previous value of n, and below the new value of n, into s, leaving one element out. Now s is 99% full. Bring in the third open set, advancing n by a factor of 100, or more, until the open set is 99.92% full. Fold these elements into s, with one missing. Now s is 99.9% full. Repeat for all open sets in the countable base. Now s is open, and cannot be the union of open sets, as it is missing something from every base set.
Hmm - maybe the previous proof was simpler after all. Oh well; it's just a contrived topology that you'll never see again, so don't worry about it.