Assume you have a proper filter $F$ that avoids both $b$ and $\neg b$. Then, you could consider the filter generated by $F\cup{b}$ - which is to say the smallest filter $F'$ containing $F$ and $b$.
Since $F$ was a proper filter it follows that $0\not\in F$.
If $0\in F'$, then this means that there is some $f\in F$ F'$such that$b\wedge f = 0$. Now,$\neg b=0\vee\neg b=(b\wedge f)\vee\neg b=(b\vee\neg b)\wedge(f\vee\neg b)=1\wedge(f\vee\neg b)=f\vee\neg b$. Thus$f≤\neg b$, which means that$\neg b\in F'$. Since$\neg b\in F'$, either$\neg b\in F$or$\neg b$may be acquired by meets and upwards closures from$F\cup\{b\}$. Say$b\wedge f≤\neg b$for some$f\in F$. Then$b\wedge f= b\wedge f\wedge\neg b = b\wedge\neg b\wedge f = 0\wedge f = 0$for an$f\in F$and by the above argument, which we derive$\neg b\in F$. This is a contradiction, from which we can derive that$0\not\in F$. Hence,$0\not\in F'$, and thus$F'$is a proper ideal strictly containing$F$. Now use Zorn's lemma, as Robin Chapman suggested, to climb up to a maximal filter. The argument above, though, shows that a maximal filter cannot avoid a pair$b, \neg b$. 1 Assume you have a proper filter$F$that avoids both$b$and$\neg b$. Then, you could consider the filter generated by$F\cup{b}$- which is to say the smallest filter$F'$containing$F$and$b$. Since$F$was a proper filter it follows that$0\not\in F$. If$0\in F'$, then this means that there is some$f\in F$such that$b\wedge f = 0$. Now,$\neg b=0\vee\neg b=(b\wedge f)\vee\neg b=(b\vee\neg b)\wedge(f\vee\neg b)=1\wedge(f\vee\neg b)=f\vee\neg b$. Thus$f≤\neg b$, which means that$\neg b\in F$, which is a contradiction. Hence,$0\not\in F'$, and thus$F'$is a proper ideal strictly containing$F$. Now use Zorn's lemma, as Robin Chapman suggested, to climb up to a maximal filter. The argument above, though, shows that a maximal filter cannot avoid a pair$b, \neg b\$.