Complete category reduces to preordered class 
Let $\mathcal{C}$ be a category such that for every category
$\mathcal{D}$ and every functor $F:\mathcal{D}\rightarrow\mathcal{C}$,
the functor $F$ has a limit. Then $\mathcal{C}$ is a preordered class.

In order to prove this, we must show that, for every $X,Y\in\mathcal{C}$, the set $\mathcal{C}(X,Y)$ has at most one element.
Suppose there exist $X,Y\in\mathcal{C}$ such that $f,g:X\rightarrow Y$ and $f\ne g$. Let $\text{Arr}(\mathcal{C})$ be the arrow category of $\mathcal{C}$ and $\Delta_Y:\text{Arr}(\mathcal{C})\rightarrow\mathcal{C}$ be the constant functor on $Y$. By assumption, there exist $L\in\mathcal{C}$ and a family of morphisms $(p_u:L\rightarrow Y)_{u\in\text{Arr}(\mathcal{C})}$ such that
$$p_{u'}=\Delta_Y(h,k)\circ p_u=1_Y\circ p_u=p_u$$
for $(h,k):u\rightarrow u'$.
The next step is to show that the cardinality of the set
$$A:=\left\{(M,(q_u)_{u\in\text{Arr}(\mathcal{C})})\ |\ \forall u,\,q_u:M\rightarrow Y\land\forall(u\rightarrow u'), q_u=q_{u'}\right\}$$
is $\geq2^{\text{Arr}(\mathcal{C})}$. ($A$ is supposed to be the set of cones on $\Delta_Y$.)
But: (i) Why should $A$ be a set? (ii) Even if it is a set, how can I  show that $|A|\geq2^{\text{Arr}(\mathcal{C})}$? I already know that I can use $f$ and $g$ to construct two cones on $\Delta_Y$.
Edit:
Is every pair $(M,q)\in\{X\}\times\{f,g\}^{\text{Arr}(\mathcal{C})}$ a cone on $\Delta_Y$? Take $q\in\{f,g\}^{\text{Arr}(\mathcal{C})}$. Let $(h,k):u\rightarrow u'$. If $q_u=f$ and $q_{u'}=g$, then we cannot have
$$q_u=\Delta_Y(h,k)\circ q_{u'}$$
because that would imply $f=g$. So arbitrary $(M,q)\in\{X\}\times\{f,g\}^{\text{Arr}(\mathcal{C})}$ cannot be a cone..
Edit 2:
Can some please explain why there are at least $2^{\text{Arr}(\mathcal{C})}$ distinct cones on $\Delta_Y$? How can I use $f,g:X\rightarrow Y$ and $(p_u:L\rightarrow Y)_{u\in\text{Arr}(\mathcal{C})}$ to construct cones on $\Delta_Y$? (I know that the constant classes of morphisms $(f)_{u\in\text{Arr}(\mathcal{C})}$ and $(g)_{u\in\text{Arr}(\mathcal{C})}$ together with $X$ constitute two cones.)
 A: I think part of your confusion is that you are trying to use the arrow category $\operatorname{Arr}(\mathcal{C})$, while you should just take the $\operatorname{Arr}(\mathcal{C})$-fold product of $Y$. Let's denote by $\operatorname{arr}(\mathcal{C})$ the discrete category of arrows in $\mathcal{C}$. So it is just the collection of arrows in $\mathcal{C}$. The product that we are interested in is
$$
\prod_{u \in \operatorname{arr}(\mathcal{C})} Y.
$$
Let's assume for a moment that $\mathcal{C}$ is small. So $\operatorname{arr}(\mathcal{C})$ is a set and its powerset $2^{\operatorname{arr}(\mathcal{C})}$ makes sense. For any $U \subseteq \operatorname{arr}(\mathcal{C})$ we can define a cone $(X, p_u)_{u \in \operatorname{arr}(\mathcal{C})}$ as follows:
$$
p_u = \begin{cases}
f & \text{if } u \in U \\
g & \text{else}
\end{cases}
$$
As per your notation $f$ and $g$ are the distinct arrows $X \to Y$. So there is an injection from subsets of $\operatorname{arr}(\mathcal{C})$ to the collection of cones on our product diagram. By the universal property there is a bijective correspondence between these cones and arrows $X \to \prod_{u \in \operatorname{arr}(\mathcal{C})} Y$. We thus find injections
$$
2^{\operatorname{arr}(\mathcal{C})} \hookrightarrow
\operatorname{Hom}_\mathcal{C}(X, \prod_{u \in \operatorname{arr}(\mathcal{C})} Y) \subseteq
\operatorname{arr}(\mathcal{C}),
$$
contradicting Cantor's theorem on the cardinality of a powerset.
This trick still works even if $\mathcal{C}$ is not small. One way to easily fix it is to just use universes. So $\operatorname{arr}(\mathcal{C})$ is a set in some bigger universe. If you are uncomfortable with such foundational assumptions, we can just unfold Cantor's argument.
So now we no longer assume $\mathcal{C}$ is small (not even in some bigger universe). Denote by $\pi_u: \prod_{u \in \operatorname{arr}(\mathcal{C})} Y \to Y$ the projection on the $u$ coordinate. Define a cone $(X, q_u)_{u \in \operatorname{arr}(\mathcal{C})}$ on our product diagram as follows. For $u \not \in \operatorname{Hom}_\mathcal{C}(X, \prod_{u \in \operatorname{arr}(\mathcal{C})} Y)$ we set $q_u = f$ (we don't really care about this part of the cone). For $u \in \operatorname{Hom}_\mathcal{C}(X, \prod_{u \in \operatorname{arr}(\mathcal{C})} Y)$ we set
$$
q_u = \begin{cases}
f & \text{if } \pi_u u = g \\
g & \text{else}
\end{cases}
$$
By the universal property $(X, q_u)_{u \in \operatorname{arr}(\mathcal{C})}$ corresponds to an arrow $v: X \to \prod_{u \in \operatorname{arr}(\mathcal{C})} Y$. That means that $q_v = \pi_v v$, which is impossible by how we constructed $q_v$. So we find our contradiction and conclude that there cannot be two parallel distinct arrows.
