Vtyjkyuk

What is the guarantee of existence of $B_n$ with underlined property?Please help me with the proof.

A local base $mathcalB$ at $x$ for a topology has the property that for every open set $O$ in that topology such that $x in O$, there is some $B in mathcalB$ such that $x in B subseteq O$.

Now for any fixed $y neq x$, $O = Xsetminus y$ is open in the co-finite topology (as its complement is $y$ which is quite finite) and contains $x$ because $x neq y$.
So we have $B_n(y)$ in the base such that $x in B_n subseteq Xsetminusy$, where the latter inclusion just says that $y notin B_n(y)$.

So taking all those $B_n(y)$ for all $y neq x$, their intersection cannot contain any $y neq x$ anymore and so the intersection is $x$ exactly.

It's IMHO a bit clearer to just start out saying that each $B_n$ in the supposed local base is of the form $Xsetminus F_n$ where $F_n$ is finite. Because all open sets that are not empty (and they contain $x$ so they're not empty) are complements of finite sets by definition.

So for $y neq x$ we have a finite subset $F_n$ (where $n$ depends on $y$) such that $x in Xsetminus F_n subseteq Xsetminus y$ by the same property of local bases. But then $y in F_n$. (e.g. use that $Xsetminus A subseteq Xsetminus B$ iff $B subseteq A$) and so the $F_n$ we collect from doing this for all $y neq x$ have the property that their union is $Y setminus x$, which is uncountable while we have a countable union of finite sets. Instant contradiction, so no such countable local base exists.