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 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.
