Noetherian Lemma Contradiction There is a noetherian version of Higman's Lemma which says

If $X$ is a noetherian poset, then $X^*$ is noetherian.

Now I was thinking, given $X=\{1\}$ we get $X^*=\{1,11,111,\dots\}$ by the concatenation of $X$. In addition, say $X=\mathbb{N}$; however $\mathbb{N}$ is not noetherian with the usual order $<$. If $X$ noetherian, how is $X^*$ noetherian when it never seems to stabilize?
 A: 
Short version: This theorem appears to be a statement about topologies, in which case the topology on $X^*$ is not supposed to be the topology induced by the subword order.

The version of Higman's Lemma I was able to find which mentioned a Nötherian condition was in a topological context, for instance, in Goubault-Lorrecq's On Noetherian spaces and verification.
On slide 38, he mentions this result, along with a note that says we require $X^*$ to be equipped with the subword topology: every finite list of open sets $\ell=(U_1,U_2,\dots U_n)$ in $X$ defines an open set: 
$$B_\ell = \{\psi: \text{there exists a subword } \zeta\subseteq\psi \text{ such that }\zeta\in U_1U_2\cdots U_n\}.$$
where the product of open sets in $X$ is the set (in $X^*$) of all concatenations of their elements. The subword topology is then the topology generated by the basis of all $B_\ell$.
Specializing to the case where $X=\{1\}$, there is only one nonempty open set, so if $\ell$ contains $k$ nonempty open sets, then $B_\ell=\{\psi: \text{there exists a subword }\zeta=1^k\}$. Of course, this simply means that $B_\ell$ contains every word in $X^*$ except those with fewer than $k$ ones.
You can see that this topology is much finer than the one induced by the subword order. It does not, for instance, have $\{11,111,1111\}$ as an open set. So if we think of Nötherian by any sort of restricted-structure-of-subobjects condition, it should be much easier than we expected for $X^*$ to have the desired restriction.

We can actually show, with our bare hands, that $X^*$ is Nötherian for the case you describe. You may find this useful for understanding the subword topology. I will leave to you the main result (which is not hard to prove, but which you should do yourself if you want to see what's happening):
Exercise: When $X=\{1\}$, any union of basis elements for $X^*$ is another basis element; moreover, it is one of the elements which was unioned together.
Going back to the presentation: On slide 13, he defines a Nötherian subspace as one for which every open set is compact. 
Specializing again to the set $X=\{1\}$, consider any open set in $X^*$: by the exercise this is a basic open set $B_\ell$. Now consider any open cover of $B_\ell$; by the exercise, this is a basic cover (consists of basic sets), and so again by the exercise, there is a single element $B$ in the cover which covers $B_\ell$ by itself. 
Therefore, this cover does have an finite subcover, namely $\{B\}$, and hence $X^*$ is Nötherian.
