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
 A: 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.
A: A few corrections:
First, if $U$ is open in the cofinite topology then $\Bbb R\setminus U$ is either finite or it is $\Bbb R$ itself (if $U = \emptyset$). There's no simpler description of a basis for the topology.
Second, it's not true that $U = \bigcup\limits_{x \in U} \{B_\epsilon(x)|\epsilon > 0\}$. For one thing, you probably mean $B_\epsilon(x) \setminus \{x\}$ not the entire ball $B_\epsilon(x)$, because $x\in B_\epsilon(x)$. For another, if $x\in U, y\notin U$, then $y \in B_\epsilon(x) \setminus \{x\}$ for some $\epsilon$.
An approach that works: Suppose $U$ is open in the cofinite topology. If $U=\Bbb R$ then $U$ is open in the usual topology, so suppose not. Then $\Bbb R \setminus U$ is finite, so in the usual topology it's closed, hence its complement $\Bbb R \setminus (\Bbb R \setminus U) = U$ is open.
To show that the two topologies are different, you need to find an open set in the usual topology that isn't open in the cofinite topology. Consider $U = \Bbb R \setminus \Bbb N$. Thus, the usual topology is strictly finer.
A: Your notation $\bigcup_{x\in U}\{B_{\epsilon}(x) \mid \epsilon >0 \}$ doesn't make sense - what is $\epsilon$ here? Instead, you are almost right - for each $x \in U$ you must find one $\epsilon_x$ (possibly depending on $x$) so that $B_{\epsilon_x} \subset U$. Since the complement of $U$ is a finite set, you can choose $\epsilon_x$ as $\min \{|x-z| \mid z\in \mathbb R \setminus U \}$. The complement of U being a finite set forces this $\epsilon_x$ to be strictly positive (the min of a finite set of strictly positive numbers).
