"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.
The way I understand minimal and maximal is the following.
Consider the following collection of sets:
{{1,2},{2,4},{1,6,7},{1,2,4,5,6},{8}}
The minimal members of this collection of sets is: {1,2},{2,4},{1,6,7},{8}
Meaning there are no other sets in the collection that are proper subsets of the set.
Minimal Set Example 1: in the set {1,2} this case holds true because the only proper subsets possible are {1} and {2}. Those sets {{1},{2}) are nowhere to be found in the collection. Therefore we include {1,2} as a part of the minimal collection of sets.
Minimal Set Example 2: in the set {1,6,7} this case holds true because the only proper subsets possible are {1},{6},{7},{1,6},{1,7}, and {6,7}. Those proper subsets ({1},{6},{7},{1,6},{1,7}, and {6,7}) are nowhere to be found in the collection. Therefore we include {1,6,7} as a part of the minimal collection of sets.
The maximal members of this collection of sets is: {1,6,7},{1,2,4,5,6},{8}
Meaning none of these sets are proper subsets of other sets in the collection. NOTICE: The maximal collection is not mutually exclusive of the minimum collection. See {1,6,7}.
Maximal Set Example 1: in the set {8} this case holds true because this set is not found to be a subset of any of the other sets in the collection. Meaning 8 is not found in any of the other sets in the collection. Therefore we include {8} as a part of the maximal collection of sets.
Maximal Set Example 2: in the set {1,6,7} this case holds true because this set is not found to be a subset of any of the other sets in the collection. Meaning all of the numbers of the set {1,6,7} are not found together as a part of another set in the collection. Therefore we include {1,6,7} as a part of the maximal collection of sets.
Best Answer
Edit: In response to the comments a proof using Zorn's lemma is also included in this answer. Notice that if $X$ is either well-ordered or does not have a least element there is nothing more to prove. The two sets are $X$ and $\varnothing$. Suppose that $X$ has a least element $x_{0}$ but $X$ is not well-ordered.
Define $W_{0} = \{ x_{0} \} $. Suppose that $\beta$ is an ordinal and for all ordinals $\alpha < \beta$ the set $W_{\alpha}$ has been defined and is well-ordered. Furthermore if $\alpha_{0} \leq \alpha_{1} < \beta$, then $W_{\alpha_{0}} \subseteq W_{\alpha_{1}}$.
If $\beta$ is a limit ordinal define $W_{\beta} = \cup \{ W_{\alpha} \colon \alpha < \beta \} $.
If $\beta = \alpha + 1$ there are two cases.
Case 1. The set $X \smallsetminus W_{\alpha}$ has a least element, say $x_{\beta}$. In this case define $W_{\beta} = W_{\alpha + 1} \cup \{ x_{\beta} \} $.
Case 2. The set $X \smallsetminus W_{\alpha}$ does not have a least element. In this case stop the construction. Define $A = W_{\alpha}$ and $B = X \smallsetminus W_{\alpha}$.
Since $X$ is a set, eventually you will run out of elements that can be well-ordered as in case 1. This puts you in case 2.
Proof using Zorn's lemma.
As above suppose that $X$ has a least element but $X$ is not well ordered. Define $$\mathcal{B} = \{ B \subseteq X \colon B \text{ does not have a least element} \text{.} \} $$ Since $X$ is not well-ordered, the set $\mathcal{B}$ is not empty. Suppose that $\mathcal{T} \subset \mathcal{B}$ is nonempty and totally ordered by inclusion. Select $t \in \cup \mathcal{T}$. For some $T \in \mathcal{T}$ we have $t \in T$. Since $T \in \mathcal{B}$, there is a $t^{*} \in T$ which satisfies $t^{*} < t$. But then $t$ is not the least element in $T$, which in turn means $t$ is not the least element in $\cup \mathcal{T}$. Thus $\cup \mathcal{T} \in \mathcal{B}$. Using Zorn's lemma, there is a maximal $B^{*} \in \mathcal{B}$.
Suppose that $X \smallsetminus B^{*}$ is not totally ordered. Then there exist $y_{0}, y_{1} \in X \smallsetminus B^{*}$ such that $y_{0} \not \leq y_{1}$ and $y_{1} \not \leq y_{0}$. Notice that $B^{*} \cup \{ y_{0}, y_{1} \}$ does not have a least element. This contradicts the maximality of $B^{*}$.
Suppose that $X \smallsetminus B^{*}$ is not well-ordered. Then there exists a nonempty $E \subseteq X \smallsetminus B^{*}$ which does not have a least element. Notice that $B^{*} \cup E$ does not have a least element. This contradicts the maximality of $B^{*}$.
Thus $X \smallsetminus B^{*}$ and $B^{*}$ are the desired sets.