FALSE: A continuous functor on a complete category necessarily has a left adjoint

Regard the ordinals as a definable-class thin category under their usual order and take its opposite. Let $A$ be the set of object ordinals of a small diagram. By [L4], $\bigcup A$ is an ordinal; each $\alpha \in A$ satisfies $\alpha \subseteq \bigcup A$, so $\alpha \le \bigcup A$ by the inclusion criterion in [L4], and if $\alpha \subseteq \beta$ for every $\alpha \in A$ then $\bigcup A\subseteq \beta$. Hence $\bigcup A$ is the least upper bound of $A$ in $\mathbf{O}\mathbf{r}\mathbf{d}$, that is, a greatest lower bound and so a limit in the opposite category; for the empty diagram the union is $0$. Thus ${\mathbf{O}\mathbf{r}\mathbf{d}}^{\mathrm{o}\mathrm{p}}$ is complete in the small-diagram sense of [L1].

Suppose $U$ had a left adjoint $F$, and put $\beta =F(\ast )$. Both ${\mathbf{O}\mathbf{r}\mathbf{d}}^{\mathrm{o}\mathrm{p}}$ and $\mathbf{1}$ are locally small, being thin, so [L5] applies and gives ${\mathrm{Hom}}_{{\mathbf{O}\mathbf{r}\mathbf{d}}^{\mathrm{o}\mathrm{p}}}(\beta ,\alpha )\cong {\mathrm{Hom}}_{\mathbf{1}}(\ast ,U(\alpha ))$, which is a singleton; hence the left side is nonempty for every ordinal $\alpha$. In the opposite ordinal order this says $\alpha \le \beta$ for every ordinal $\alpha$.

The unique functor $U:{\mathbf{O}\mathbf{r}\mathbf{d}}^{\mathrm{o}\mathrm{p}}\to \mathbf{1}$ preserves every small limit, because every cone in the terminal category is limiting. It is therefore continuous by [L2].

By [L4] the successor ${\beta }^{+}=\beta \cup \{\beta \}$ is an ordinal with $\beta \in {\beta }^{+}$, so $\beta \subseteq {\beta }^{+}$, while $\beta \mathrm{\ne }{\beta }^{+}$ because $\beta \mathrm{\notin }\beta$; hence $\beta <{\beta }^{+}$, contradicting the conclusion of step 1.2 that every ordinal is at most $\beta$. Therefore no such left adjoint exists, even though the source is complete and the functor is continuous.