What is the guarantee of existence of $B_n$ with underlined property? 
What is the guarantee of existence of $B_n$ with underlined property?Please help me with the proof.
 A: A local base $\mathcal{B}$ 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 $B \in \mathcal{B}$ such that $x \in B \subseteq O$. 
Now for any fixed $y \neq x$, $O = X\setminus \{y\}$ 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 $x \in B_n \subseteq X\setminus\{y\}$, 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 $X\setminus F_n$ 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 $x \in X\setminus F_n \subseteq X\setminus \{y\}$ by the same property of local bases. But then $y \in F_n$. (e.g. use that $X\setminus A \subseteq X\setminus B$ 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.  
