How to show the usual topology is finer than co-finite topology on $\mathbb{R}$ I have solved a bunch of problems where the basis is used to quickly deduce which topology is finer than which.
However, I do not know the basis of co-finite topology.

What is the straight forward approach to compare the two topologies?

Proof attempt:
Let $U$ be an open set in $(\mathbb{R},\tau_{co-finite})$, then $\mathbb{R}\backslash U$ is finite.
We want to show that $U$ is in $((\mathbb{R},\tau_{usual})$
Let $x \in U$, then $U = \bigcup\limits_{x \in U} \{B_\epsilon(x)|\epsilon > 0\}$
Therefore $U$ is in $\tau_{usual}$
Let me know if there is some (or a lot) problem with the attempt, it's my first try

Since the open intervals form a basis of $\tau_\textsf{usual}$, it suffices to express $U$ as a union of open intervals. Indeed, since $\mathbb R \setminus U$ is finite, we know that for some $n \in \mathbb N$ and for some $x_1, \ldots, x_n \in \mathbb R$ such that $x_1 < \cdots < x_n$, we have that:
$$
\mathbb R \setminus U = \{x_1, \ldots, x_n\} \iff
U = (-\infty, x_1) \cup (x_1, x_2) \cup \cdots \cup (x_{n-1}, x_n) \cup (x_n, \infty)
$$
as desired.
