Fortissimo space

Choose an uncountable set $X$ and a point $a\in X$. Every point other than $a$ is isolated, and a set containing $a$ is open precisely when its complement is countable. Equivalently, the closed sets are all subsets containing $a$ together with all countable subsets. Arbitrary intersections and finite unions preserve this latter condition, so it defines a topology.

Every convergent sequence in this space is eventually equal to its limit. Indeed the set of its values different from its limit is countable and closed; its complement is an open neighbourhood of the limit. Eventual membership in this complement forces eventual equality. Nevertheless $a$ is not isolated, because its singleton has uncountable complement. In particular sequence convergence alone does not determine this topology.

A real-valued function on this space is continuous exactly when $\{x:f(x)\ne f(a)\}$ is countable. Sufficiency follows by considering the preimages of closed subsets of the real line: each either contains $a$ or is a subset of that countable exceptional set. For necessity, each set $\{x:|f(x)-f(a)|\ge 1/n\}$ is closed and excludes $a$, hence is countable; their countable union is the exceptional set.