Trouble understanding this proof about the minimum number of states of a deterministic finite automaton
Let $L$ be an arbitrary language. We say that two words $x,y$ are equivalent modulo $L$, written $x \stackrel{L}{\sim} y$, if for all words $z$, either both $xz,yz$ are in $L$, or both are not in $L$.

Lemma 1. Suppose that $x_1,\ldots,x_n$ are pairwise inequivalent modulo $L$. Then any DFA for $L$ must contain at least $n$ states.

Proof. Consider a DFA accepting $L$. Let $q_i$ be the state that the DFA reaches after reading $x_i$. I claim that $q_1,\ldots,q_n$ are all distinct. Indeed, consider any two states $q_i,q_j$. Since $x_i,x_j$ are inequivalent modulo $L$, there exists a word $z$ such that $x_iz \in L$ and $x_jz \notin L$, or vice versa; without loss of generality, assume the former. Since $x_iz \in L$, $\delta(q_i,z)$ must be an accepting state, whereas since $x_jz \notin L$, $\delta(q_j,z)$ must not be an accepting state. This shows that $q_i \neq q_j$. $\quad\square$

This is the method which is used in Proposition 5 that you mention.

What is the best lower bound that Lemma 1 can give us? Once you notice that $\stackrel L \sim$ is an equivalent relation, it is not hard to check that the maximum $n$ that you can choose is the number of equivalence classes of $\stackrel L \sim$:

Corollary 2. Any DFA for $L$ must contain at least $n$ states, where $n$ is the number of equivalence classes of $\stackrel L \sim$. In particular, if $\stackrel L \sim$ has infinitely many equivalence classes, then $L$ is not regular.

It is natural to ask whether this method is optimal:


If $L$ is not regular, can we always prove this using Corollary 2?
If $L$ is regular, is there always a DFA having $n$ states?


The positive answer to both questions is the main result of Myhill–Nerode theory, which you might learn about in the future.