Nesthood has Finite Character/Proof 2

Proof
By definition, a nest is a class on which $\subseteq$ is a total ordering.

Sufficient Condition
Let $x$ be a nest.

Let $y \subseteq x$.

From Restriction of Total Ordering is Total Ordering it follows that $y$ is also a nest.

This holds in particular if $y$ is a finite set.

Hence the result.

Necessary Condition
Suppose that every finite subset of $x$ is a nest.

We have that Subset Relation is Ordering.

It remains to show that $\subseteq$ is total.

Let $y, y' \in x$ be arbitrary.

Since $\set {y, y'}$ is a finite subset of $x$, it is a nest.

Therefore, either $y \subseteq y'$ or $y' \subseteq y$.

Since $y, y'$ were arbitrary, it follows that $x$ is also a nest.