The problem (taken from Kaplansky's Set theory and metric spaces, p. 40) is as follows
Let $L$ be a lattice in which every chain has a least upper bound and a greatest lower bound. Prove that $L$ is complete.
Using the fact that $L$ is a lattice in which every chain has an upper bound I was able to prove that $L$ has a unique maximal element (with Zorn's lemma). With this I can prove that an empty set and any subset of $L$ containing its maximal element have a least upper bound.
But I get stuck trying to prove that a non-empty subset of $L$ not containing the maximal element has a least upper bound. How can I do this? Note that I can not use the axiom of choice here, only Zorn's Lemma.
I'll show that Kaplansky's assumptions imply that every subset $X$ of your lattice $L$ has a greatest lower bound.
Step 1: Let $Y$ be the set of all lower bounds of $X$. I claim that, if $C$ is a chain in $Y$, then $C$ has an upper bound (in fact a least upper bound) in $Y$. By assumption, $C$ has a least upper bound $b\in L$. Because $b$ is least, every element of $X$, being an upper bound for $C$, is $\geq b$. This means $b\in Y$, as required.
Step 2: In view of Step 1, we can apply Zorn's Lemma to conclude that $Y$ has a maximal element $m$.
Step 3: I claim that $m$ is $\geq$ all elements of $Y$. To see this, consider any $y\in Y$. Each element of $X$ is $\geq$ both $m$ and $y$ (since they're both in $Y$) and therefore also $\geq m\lor y$. So $m\lor y$ is an element of $Y$ that is $\geq m$. By maximality of $m$, we conclude that $m\lor y=m$, which is equivalent to $y\leq m$.
So $m$ is the largest element of the set $Y$ of lower bounds of $X$, i.e., $m$ is the greatest lower bound of $X$.