Is this proof of a topology not having a countable base correct? Let $X = \mathbb{R}$ and consider the topology $\tau = \{\emptyset\}\cup\{A\subseteq X: A^c = X-A\text{ is finite}\}$.  I want to show that $\tau$ does not have a countable base. Here is my attempt. 
Let $\mathbb{B}$ be a base for $\tau$. Note that any interval $(-\infty,r)\cup (r, \infty)$ where $r$ is irrational is open. So there must be a $B\in \mathbb{B}$ such that $r\notin B$. This $B$ is 'missing' only finitely many real numbers. We can find another irrational $r'$ that is different from $r$ and all the other things $B$ is missing. Then we can find $B'\in \mathbb{B}$ such that $r'\notin B'$. Again this $B'$ is missing only finitely many numbers, so we can repeat this process again. We end up with a different element in $\mathbb{B}$ for every number in the set of irrationals minus a union of finite sets. So we have found an uncountable number of elements in $\mathbb{B}$. Is this correct?
 A: Let $\mathbb B$ be a base of the cofinite topology on an uncountable set $X$. 
Then for every $x\in X$ we can find some cofinite $B\in\mathbb B$ with $x\notin B$. 
For this just let $y\in X$ with $y\neq x$ and choose some $B\in\mathbb B$ with $y\in B\subseteq X-\{x\}$.
This implies that the intersection of all elements of $\mathbb B$ must be empty. 
However, if $\mathbb B$ is a countable collection of cofinite sets then the complement of the intersection is a countable set hence does not coincide with $X$.
This tells us directly that base $\mathbb B$ cannot be countable.
A: The idea of your proof is correct. However your proof relies on the induction for infinite objects. Your proof is inductive, that is, in each step of your proof relies on previous steps, and you have to generate uncountable collection 
of open sets chosen from $\mathbb{B}$.
Here is a point the problem comes. Assume that you have chosen $B_1$, $B_2$, $\cdots$ by your construction. The proof is not ended because you need to choose more open sets. How to continue? You should mention about the limit case.
You can make your proof rigorously by introducing transfinite induction. However there is another proof that does not require the transfinite induction:
For each countable collection $\mathbb{B}$ of open sets in $\mathbb{R}$ with your topology, consider $X=\bigcup\{\mathbb{R}\setminus B : B\in\mathbb{B}\}$. $X$ is a countable union of countable sets, so $X$ is countable. Hence we can find some $r\in\mathbb{R}\setminus X$. 
If $\mathbb{B}$ is really a basis, then $\mathbb{R}\setminus \{r\}$ is a union of some elements of $\mathbb{B}$. Equivalently, $\{r\}$ is an intersection of sets $X\setminus B_i$, $i=1,2,\cdots$, for some $B_i\in\mathbb{B}$. However $\bigcap_i (X\setminus B_i)\subseteq X$, whereas $r\notin X$, so $\{r\}$ can't be an intersection of sets of the form $X\setminus B_i$.
A: The easiest way is the direct proof (no iterations etc.) as follows:
Suppose that $(B_n)_{n \in \mathbb{N}}$ is a countable base of non-empty sets.
Then for each $n$ the set $F_n = \mathbb{R}\setminus B_n$ is finite.
Then $F = \cup_n F_n$ is at most countable and the reals are uncountable so pick $x \in \mathbb{R}\setminus F$. 
Then $X \setminus \{x\}$ is open but no $B_n$ is a subset of it, as 
$B_n \subset X \setminus \{x\}$ iff $x \in X \setminus B_n = F_n$. But $x$ is in none of the $F_n$. So we cannot write some open subset as the union of base elements, contradiction. A countable base thus does not exist.
