Suppose that $X$ is a set of ordinals. First note that by the axiom of union we have that $\bigcup X$ is a set as well.
First we want to show that $\bigcup X$ is transitive, that is $x\in\bigcup X$ then $x\subseteq X$. Suppose that $x\in\bigcup X$, then there is some $\alpha\in X$ such that $x\in\alpha$. Since $\alpha$ is an ordinal it is a transitive set and therefore $x\subseteq\alpha$, and since $\alpha\subseteq\bigcup X$ we have $x\subseteq\bigcup X$ as wanted.
Now we want to show that it is linearly ordered by $\in$, so take $x,y\in\bigcup X$. Then there are $\alpha,\beta\in X$ such that $x\in\alpha$ and $y\in\beta$. By Lemma 2.11 (p. 19) we know that $\alpha\in\beta$ or $\beta\in\alpha$ or $\alpha=\beta$. Either way we may assume without loss of generality that $x,y\in\alpha$. Since $\alpha$ is linearly ordered by $\in$ we have that $x\in y$ or $y\in x$ or $x=y$ as wanted.
Lastly, to show that every non-empty set has a least element we can resort to the axiom of regularity which says that $\in$ is well-founded. To see that directly we can also argue as following:
Let $A\subseteq\bigcup X$ be non-empty. There is some $\alpha\in X$ such that $A\cap\alpha\neq\varnothing$. Since $\alpha$ is well-ordered we have that $\alpha\cap A$ has a minimal element, call it $x$. Suppose that $x$ was not minimal in $A$, then there is some $y\in A$ such that $y\in x$, however $x\in\alpha$ therefore $x\subseteq\alpha$, so $y\in\alpha$ and therefore $y\in A\cap\alpha$ which contradicts the fact that $x$ was the minimal element of $A\cap\alpha$.