This question is inspired by this question. Consider the following result:
Let $(L,\leq)$ be a chain-complete lattice. Then $(L,\leq)$ is a complete lattice.
Can this result be proven without using the axiom of choice?
For completeness, here is a proof that the result holds under the axiom of choice: Let $S\subseteq L$. We want to show that $S$ has an upper bound. Since $\emptyset$ is a chain, we can assume $S\neq\emptyset$. Let $\preceq$ be a well ordering of $S$. Fix $s\in S$. Define a transfinite sequence in the following way: Let $s_0=s$. For a successor ordinal $\alpha+1$ we let $s_{\alpha+1}=s_\alpha\vee m$ with $m$ being the $\preceq$-smallest element in $S$ that is not smaller than $s_\alpha$. For a limit ordinal $\beta$, let $s_\alpha=\sup_{\beta<\alpha}s_\alpha$ which exists as the supremum of a chain. The range of this sequence is clearly a chain and its supremum is the supremum of $S$. That $S$ has an infimum follows in the usual way by taking the infimum to be the supremum of all lower bounds.