Okay, I cheated. This is based on these notes by William Weiss.
Let $A$ be a transitive set of transitive sets. We want to prove that $A$ is totally ordered with respect to $\epsilon$. I will assume the Axiom of Regularity.
Axiom of Regularity. For every $A$, $A\neq\emptyset$, there exists $x\in A$ such that $x\cap A = \emptyset$.
First we prove $\epsilon$ is a total order on $A$. If $A$ is empty, there is nothing to do. So assume $A$ is nonempty.
Claim 1. If $a,b\in A$, and $a\subseteq b$, then either $a=b$ or $a\in b$.
Let $S = \{ b\in A\mid \text{there exists }a\in A\text{ such that }a\subseteq b\text{ and }a\neq b, a\notin b\}.$ We want to show that $S$ is empty. If $S$ is not empty, then by the Axiom of Regularity there exists $b\in S$ such that $b\cap S=\emptyset$. Let $a$ be a witness to the fact that $b\in S$ (that is, $a\subseteq b$, $a\neq b$, and $a\notin b$.
Then $b-a$ is nonempty, so by the Axiom of Regularity there exists $x\in b-a$ such that $x\cap (b-a)=\emptyset$. Since $x\in b$ and $b$ is transitive, $x\subseteq b$; since $x\cap(b-a)=\emptyset$, then we must also have $x\subseteq a$. Since $x\in b$ but $a\notin b$, then $x\neq a$. Therefore, $a-x\neq \emptyset$. Again by Regularity there exists $y\in a-x$ such that $y\cap (a-x)=\emptyset$. Since $a$ is transitive and $y\in a$, $y\subseteq a$, and as above this implies that $y\subseteq x$ as well, since $y\cap(a-x)=\emptyset$.
Since $x\in b$, and $b\in A$, then $x\in A$. Moreover, since $b\cap S=\emptyset$, then $x\notin S$, so since $y\subseteq x$, it follows that either $y=x$ or $y\in x$. But $x\in b-a$, and $y\in a-x$, so we cannot have $y=x$, hence we conclude that $y\in x$. But $y\in a-x$, which means $y\notin x$. This contradiction arises from the assumption that $S\neq \emptyset$, so $S=\emptyset$.
Therefore, for any $a,b\in A$, if $a\subseteq b$, then either $a=b$ or $a\in b$.
Claim 2. If $a,b\in A$ and $b\not\subseteq a$, then $a\in b$.
Suppose that $b\not\subseteq a$. Using Foundation, let $x\in b-a$ such that $x\cap (b-a)=\emptyset$. Since $x\in b$ and $b$ is transitive, $x\subseteq b$, hence $x\subseteq a$ as well. By Claim 1, either $x=a$ or $x\in a$; the latter is impossible since $x\in b-a$, so $x=a$. Therefore, $a\in b$.
Putting the two together: let $A$ be a transitive set of transitive sets, and let $a,b\in A$. If $b\subseteq a$, then either $b=a$ or $b\in a$, by Claim 1; if $b\not\subseteq a$, then $a\in b$ by Claim 2. Either way, given any $a,b\in A$, we have that either $a\in b$, $a=b$, or $b\in a$, proving that $\epsilon$ is a strict total order on $A$.