**Are $(\text{Cov}(X)/\simeq, \leq_q)$ and
$(\text{PropCov}(X),\leq)$ isomorphic as posets?**

If $X$ is finite, then $(\text{Cov}(X)/\simeq, \leq_q)$
and $(\text{PropCov}(X),\leq)$
are isomorphic, but if $X$ is infinite they are not.

To see why they are isomorphic when $X$ is finite it is enough to
note that, in the case of finite $X$, each $\simeq$-class
of covers has the property that all members share
the same maximal elements. Moreover, this common set
of maximal elements constitutes a proper cover,
and it is the unique proper cover in this $\simeq$-class.
This bijection between $\simeq$-classes of covers
and proper covers preserves and reflects order, hence establishes
an isomorphism between $(\text{Cov}(X)/\simeq, \leq_q)$ and
$(\text{PropCov}(X),\leq)$.

To see why $(\text{Cov}(X)/\simeq, \leq_q)$
and $(\text{PropCov}(X),\leq)$
are not isomorphic whene $X$ is infinite,
it will suffice to point out that
$(\text{Cov}(X)/\simeq, \leq_q)$ is a lattice
while $(\text{PropCov}(X),\leq)$ is not.

I first argue the easy part:
$(\text{Cov}(X)/\simeq, \leq_q)$
is a lattice.

Consider the following closure operator
on the set of covers:
given $B\in \text{Cov}(X)$, let $B^* = \bigcup_{A\leq B} A$.
It is easy to see that $B^*$ is a cover of $X$, and that
an alternative definition for it is
$$
B^* = (\bigcup_{b\in B} {\mathcal P}(b))\setminus \{\emptyset\}.
$$
Moreover, $A\leq B$ holds iff $A\subseteq B^*$ iff $A^*\subseteq B^*$,
so $A\simeq B$ iff $A^*=B^*$.

Now, a subset $Z\subseteq {\mathcal P}(X)$ is of the form $Z=B^*$
iff (i) $\emptyset\notin Z$ and (ii) $Z\cup\{\emptyset\}$
is an order ideal of ${\mathcal P}(X)$ which contains
all the singleton subsets of $X$. This is enough to show that
$(\text{Cov}(X)/\simeq, \leq_q)$ is isomorphic to the
interval of the lattice of order ideals of ${\mathcal P}(X)$
above the order ideal consisting of singleton sets.
This is a distributive, algebraic lattice.

Now the less-easy part. I argue that
$(\text{PropCov}(X),\leq)$ is not a lattice.
In fact, I will write out only the case
$X=\omega$, but afterwards I will explain how to extend the argument to
any superset of $\omega$.

Let's name some subsets of $\omega$:

$E$: the set $\{0,2,4,\ldots\}$
of all even natural numbers.

$E_n$: the set $\{0,2,4,\ldots,2n\}$
of even numbers up to $2n$.

$E_n^*$: the set $\{0,2,4,\ldots,2n,2n+1\}$
of even numbers up to $2n$
along with one odd number $2n+1$.

Now let's name some proper covers of $\omega$:

$C_E$: the proper cover consisting of $E$ and all odd singletons.

$C_{E^*}$: The proper cover consisting of all $E_n^*$'s.

$C_n$: The proper cover including $E_n$ and all singletons
of elements from $\omega \setminus E_n$.

It is not hard to see that

(1) $C_E$ and $C_{E^*}$ are incomparable under $\leq$.

(2) $C_n\leq C_E$ and $C_n\leq C_{E^*}$ for all $n$.

My goal is to show that $C_E$ and $C_{E^*}$ do not have a
meet (= greatest lower bound) in $(\text{PropCov}(X),\leq)$.

Suppose that $D$ is a proper cover of $\omega$
that is a candidate for the meet
of $C_E$ and $C_{E^*}$ in $(\text{PropCov}(X),\leq)$.
Then $D\leq C_E$, $D\leq C_{E^*}$, and $D$ is above all other
common lower bounds of
$C_E$ and $C_{E^*}$, for example $C_n\leq D$ for all $n$.

Since $E_n=\{0,2,4,\ldots,2n\}\in C_n$,
there must be a set $d_n\in D$ such that $E_n\subseteq d_n$.
Since $D\leq C_E$, it follows that $d_n$ consists of even
integers.
Since $D\leq C_{E^*}$, it follows that $d_n$ is a finite set.
Thus $d_n$ is properly contained in some
$E_m=\{0,2,\ldots,2m\}$ for sufficiently
large $m$. But now repeat this argument to find
some $d_m\in D$ containing $E_m$. We now have
$d_n\subsetneq E_m \subseteq d_m$, contradicting the
properness of $D$.
\\\\\

Finally, to extend the
proof that $(\text{PropCov}(\omega),\leq)$ is not a lattice
to a proof that $(\text{PropCov}(X),\leq)$ is not a lattice,
for any superset $X\supseteq \omega$, repeat the argument
above, but extend all the covers of $\omega$ which were used
in the argument to covers
of $X$ by adding to those covers
all singletons $\{x\}$, $x\in X\setminus\omega$.