Let $X=Y:=\mathbb R$ and define the two topological spaces $(X,\tau_{1})$ and $(Y,\tau_{2})$ where $\tau_{2}$ is the topology generated by the euclidean metric on $\mathbb R$ and $\tau_{1}:=\{\mathbb R, \varnothing\}\cup\{\mathbb R\setminus A:A \operatorname{ countable}\}$
Background: I want to show that any function $F:X\to Y$ is sequentially continuous
and I have come to the conclusion that in case of convergence (say, sequence $(x_{n})_{n}$) in $\tau_{1}$ that after a particular $N$ the sequence $(x_{n})_{n}$ is constant. But I have not found a way to prove this fact. Any ideas? After this fact, sequential continuity is a breeze.
Further: How can I show that the identity is not continuous?
Suppose that $x_n \to x$. Define $U= \Bbb R \setminus \{x_n: x_n \neq x\}$. This set is open (as its complement is $\{x_n: x_n \neq x\}$, which is a subset of the terms of a sequence, so at most countable.). It also contains $x$ as by definition $x \notin \{x_n: x_n \neq x\}$.
By the definition of convergence, there is some $N \in \Bbb N$ such that
$$\forall n \ge N: x_n \in U$$
And note that $x_n \in U$ iff $x_n \notin \{x_n: x_n \neq x\}$ iff $x_n =x$.
So in fact
$$\forall n \ge N: x_n = x$$
as required.
Finally, the identity $F: (\Bbb R, \tau_1) \to (\Bbb R, \tau_2)$ is not continuous, as $(0,1) \in \tau_2$ and $F^{-1}[(0,1)] = (0,1)$ is not open in $\tau_1$, as its complement is uncountable.