Question: Let $(D,\prec) $ be a directed set. Is there a "cofinal chain" in $D $? By chain I mean a $D'\subset D $ such that for all $\lambda,\gamma \in D'$, then $\lambda\prec \gamma $ or $\gamma\prec \lambda $. By cofinal chain I mean a chain $D'$ for which, for every $\gamma\in D $, there is some $\lambda \in D'$ such that $\gamma\prec \lambda $.

Answer: 
Take $X$ to be your favorite uncountable set. Now take $D$ to be the finite subsets of $X$, ordered by inclusion.

Easily, this set is directed. But there are no chains with order type longer than $\omega$, and so no chain can have an uncountable union, and in particular, no chain is cofinal.

(If you choose $X$ wisely, then no choice is needed either. And by wisely I just mean not a countable union of finite sets.)
If such chains always existed, say under AC, directed sets would have no use in ZFC since most constructions involving directed sets could be done using such cofinal chains.

But it is not the case. For instance let $\omega\_1$ be ordered by divisibility $|$ under Hessenberg product. For $\alpha,\beta\in \omega\_1$, $\alpha,\beta | \alpha \otimes \beta\in \omega\_1$ so this ordered set is directed. The divisibility order is contained in the standard order on $\omega\_1$ so any cofinal chain in $(\omega\_1,|)$ must have cofinality $\omega\_1$. In particular, for such a chain to exist, there must be a countable ordinal $\alpha$ with infinitely many factors (take the $\omega$-th element of the chain). This cannot be as $(\omega\_1,\oplus,\otimes)$ is contained in the UFD formed by taking its Grothendieck ring.

There should be simpler and more visual counter examples...