So, the key for understand why it is a proper subset is the symbol $\prec$ compared to the symbol $\preceq$. If $(P,\preceq)$ is a poset, then $x\prec y$ means "$x\preceq y$ and $x\neq y$."
Now, in the specific poset, $(P,\preceq) = (S,\subseteq)$, where $S$ is the set of all subsets of $\{a,b,c\}$. $S$ is a set with $8$ elements. We can list $S$ explicitly: $$S=\{\emptyset,\{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\},\{a,b,c\}\}$$.
It might be easier if you don't use $A,B,C$ for elements of $S$, since it might confuse the question between the lower case letters, $a,b,c$. Instead, let's use $X$, $Y$, and $Z$ for elements of $S$.
So $Y$ covers $X$ if (1) $X\prec Y$, and (2) there is no $Z\in S$ such that $X\prec Z\prec Y$.
(1) means $X\preceq Y$ and $X\neq Y$, But $\preceq$ in this poset is just $\subseteq$. So $X\preceq Y$ means $X\subseteq Y$. So necessarily, if $Y$ covers $X$, then $X$ is a subset of $Y$ and $X\neq Y$. But that is just the the definition of "$X$ is a proper subset of $Y$."
$Y=\{a,b,c\}$ is not a cover for $X=\{a\}$ because $Z=\{a,b\}$ has the property $X\prec Z\prec Y$, which violates the definition of "$Y$ is a cover for $X$."
You name one property you need to check: antisymmetric.
You must also ensure that the relation is reflexive, and transitive.
A relation R is a partial order on a set $S$
if it is reflexive: for all $x\in S, x R x$
if it is antisymmetric, which is the property you list
And if it is transitive: for all $x, y, z \in S,\; (x R y \text{ and}\; y R z) \rightarrow x R z$
Reflexivity is easy to check here: $(1,1),(2,2),(3,3),(4, 4) \in R$, so for all $x\in S, xRx$.
In small sets like this, it's usually easy enough to check whether or not these three properties all hold.
There are really only three pair in your set for which you need to consider antisymmetry and transitivity.
Can you find counterexamples to these properties?
All you need is one counterexample to a property, and all you need is one property to fail to hold, to conclude that a relation is NOT a partial order on a set.
If no property fails, the relation is a partial order on your set.
Edit:
To address your question as to what constitutes a formal proof: Yes, you could list all pairs to show satisfaction of all the properties. It depends on your instructor, how detailed you have to be in a proof.
For example, you could state $\lnot \exists x \in S$ such that $\lnot (x R x)$, which translates to the equivalent, $\forall x \in S, (xRx)$. And do similarly for each property, without listing all pairs.
Again, it depends on the context context in which you need to prove that the relation defines a partial order on S.
Best Answer
"A partial ordered set (poset) is just a relation on a set, right?" A partially ordered set is a set with a partial order. A (lax) partial order is a relation satisfying Reflexivity, Antisymmetry, and Transitivity. "And for the covering relation, the way the author describes seems to indicate that a covering relation is very similar to the poset, except it doesn't have transitivity?" A covering relation has antisymmetry trivially, but never has reflexivity and almost never has transtivity. You should think of a covering relation as arising from a given poset. "Or is it just saying that there isn't an element between $x$ and $y$? But wouldn't that mean there is no transitivity?" y covers x when x≺y and there's nothing in between. "≺" still has transitivity because of things like "if z covers y and y covers x then x≺z (even though z doesn't cover x)", but "covers" isn't transitive (unless you're in a boring case like "$S$ has only two elements").
"Also, is the symbol $≺$ just a generalization of the different symbols used, such as $\supset$, $\supseteq$,$\le$,$<$?" Yes, it's different so you don't confuse it with any particular one of those in any particular context.