Give an example of a well ordered set $(X,\le)$ in which there exists an element $x_0$ such that there are infinitely many elements $x\in X$ such that $x\lt x_0$.
Let $X=${$A_i | i \in \mathbb N$}$\cup \mathbb N$ where $A_i=${$1,2,...,i$}. Let the relation be inclusion of sets. Here, $x_0=\mathbb N$ obviously. I feel intuitively that my $X$ is well ordered but I'm unsure how to prove it.
On
Let $S\subseteq X$ be non-empty.
If $\mathbb{N}\notin S$, then $$ S=\bigcup_{i\in I}\{A_i\} $$ for a non-empty subset $I\subseteq\mathbb{N}$. Since $\mathbb{N}$ is well-ordered, it follows that there is a minimum element $i_o\in I$. Clearly, $A_{i_0}$ is the minimum of $S$.
The same applies if $$ S=\{\mathbb{N}\}\cup\bigcup_{i\in I}\{A_i\} $$ for a non-empty subset $I\subseteq\mathbb{N}$.
If $S=\{\mathbb{N}\}$ then clearly $\mathbb{N}$ is the minimum of $S$.
I'd use cardinality for the ordering relationship. That is, $x_i \le x_j$ in your ordering when $|x_i| \le |x_j|$ using the common meaning for less-than or equal.
Effectively, your set is the natural numbers plus $\aleph_0$. The well-ordering follows directly from that.