= Solution
\b[As printed, yes: the relation is universal.] The original PDF puts the existentially quantified element in the same subset as the universally quantified element; this is not an OCR substitution. For each element of that subset, choose the element itself as witness, using reflexivity of the underlying <well-quasi-ordering>. For an empty subset the condition is vacuous. The truth value therefore never depends on the proposed target subset. On the <power set> this is a reflexive transitive relation, and every two terms of any infinite <sequence> are related:
$$
\boxed{\text{literal printed relation}=\mathcal P(X)\times\mathcal P(X),\qquad\text{hence it is a well-quasi-ordering}.}
$$
A <well-quasi-ordering> need not be antisymmetric, so universal comparability is allowed.
If the existential element is instead intended to belong to the target subset, the natural relation is the <Hoare domination preorder>
$$
S\le_H T\quad\Longleftrightarrow\quad\forall s\in S\;\exists t\in T\;(s\le_Xt).
$$
\b[For arbitrary subsets the answer to that corrected question is no.] Here is a complete counterexample, the <Rado order>. Take
$$
R=\{(m,n)\in\mathbb N^2:m<n\},\qquad
(m,n)\le_R(k,l)\quad\Longleftrightarrow\quad
\bigl(m=k\text{ and }n\le l\bigr)\ \text{or}\ n<k.
$$
This is a <partial order>. Reflexivity is immediate. For transitivity, two same-row comparisons compose ordinarily. If the first comparison crosses rows and the second stays in its row, its target first coordinate is unchanged, so the cross-row inequality persists. If the first stays in its row and the second crosses, use $n\le l<k'$; if both cross, use $n<k<l<k'$. Antisymmetry follows because comparisons across distinct rows in both directions would require $n<k<l<m<n$.
The <Rado order> is a <well-quasi-ordering>. Given an infinite <sequence> $(m_i,n_i)$, if a first coordinate $m$ repeats infinitely often, its corresponding natural-number second coordinates have a nondecreasing pair, giving a same-row comparison. Otherwise each first coordinate occurs only finitely often, so the first coordinates in every tail are unbounded. In particular some later $m_j$ exceeds $n_0$, giving $(m_0,n_0)\le_R(m_j,n_j)$.
For each $m$, take the infinite row $S_m=\{(m,n):n>m\}$. For distinct $m,k$, choose $n>\max(m,k)$. The element $(m,n)\in S_m$ is below no member $(k,l)$ of $S_k$, since its first coordinate is different and $n<k$ fails. Thus $S_m\not\le_HS_k$ for every pair of distinct indices. These subsets form an infinite <antichain>, so the <Hoare domination preorder> on $\mathcal P(R)$ is not a <well-quasi-ordering>.
If only finite subsets were intended, the answer changes again: \b[the <finite-subset lifting of a well-quasi-order> is a <well-quasi-ordering>.] Enumerate each finite subset as a finite <word>. <Higman lemma> gives an earlier <word> embedding into a later one with every letter increased, which witnesses <Hoare domination preorder> comparison of the underlying subsets. The PDF specifies no finiteness restriction, so this observation supplements rather than replaces the literal answer and the arbitrary-subset counterexample.
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