The Myhill-Nerode theorem suggests that a language $L\subseteq \Sigma^*$ is regular iff $L$ has a finte index, and this clearly implies that a regular language has a finite index. So let's prove this direction of the Myhill-Nerode theorem directly. Assume that a language $L$ is regular, and let $\mathcal{A} = \langle Q, \Sigma, q_0, \delta, F\rangle$ be a DFA for it.
The idea is to relate the state-space of $\mathcal{A}$ and the Myhill-Nerode equivalence classes. Intuitively, if reading two prefixes in $\mathcal{A}$ leads to the same state $q$, then the acceptance of the rest of the input is determined only based on $q$, and the suffix to be read. Thus, there is no separating suffix between words that lead to the same state.
To capture this intuition formally,
we define an equivalence relation, denoted $\sim_{\mathcal{A}}$ over $\Sigma^*$ as follows. For every two words $x, y \in \Sigma^*$, we say that $x \sim_{\mathcal{A}} y$ iff $\delta^*(q_0, x) = \delta^*(q_0, y)$. In words, $x$ and $y$ are equivalent iff the automaton $\mathcal{A}$ reaches the same states upon reading $x$ and $y$.
The following lemma can be proved by induction and I leave it to you.
Lemma: $\delta^*(q, u\cdot v) = \delta^*(\delta^*(q, u), v)$, for every state $q$ and words $u$ and $v$.
Note: I used $\delta^*$ to denote the extension of the transition function $\delta$ to words. Thus, $\delta^*(q, w)$ is the state that is reached from $q$ upon reading the word $w$. (this is a standard extension and can be defined by induction on $w$ -- I leave the details to you).
We next show that:
The number of equivalence classes of $\sim_{\mathcal{A}}$ equals $|Q|$.
The Myhill-Nerode equivalence relation w.r.t $L$, which I denote by $\sim_L$, is coarser than the relation $\sim_{\mathcal{A}}$:
for every two words $x, y\in \Sigma^*$, if $x\sim_{\mathcal{A}} y$ then $x\sim_L y$.
Note that $2$ implies that the index of $L$ is at most the number of the equivalence classes of $\sim_{\mathcal{A}}$, and thus by $1$ and the fact that $|Q|$ is finite, we get that $L = L(\mathcal{A})$ has a finite index.
Proof of 1 and 2:
Follows immediately by the definition of $\sim_{\mathcal{A}}$. Indeed, every state $q$ of $\mathcal{A}$ defines an equivalence class.
Assume that $x$ and $y$ are such that $x \sim_{\mathcal{A}} y$. By definition we have that $\delta^*(q_0, x) = \delta^*(q_0, y)$. Let $z\in\Sigma^*$ be a word. We need to show that $xz\in L$ iff $yz\in L$: using the above lemma and what we have so far, we get the following:
$$xz \in L$$ $$\text{iff}$$
$$\delta^*(q_0, xz) \in F$$ $$\text{iff} \ \ (\text{By the lemma})$$
$$\delta^*(\delta^*(q_0, x), z) \in F$$ $$\text{iff} \ \ (\text{By the fact that $\delta^*(q_0, x) = \delta^*(q_0, y)$})$$
$$\delta^*(\delta^*(q_0, y), z) \in F$$ $$\text{iff} \ \ (\text{By the lemma})$$
$$\delta^*(q_0, yz) \in F$$ $$\text{iff}$$
$$yz\in L$$
Thus, $x \sim_L y$ and so we are done.
