How to prove that minimal DFA is unique?
Suppose that we are given a regular language $L$ and a DFA ${\cal A} = (Q,q_0,F,\delta)$. Denote the equivalence classes of the Myhill–Nerode relation of $L$ by $S_1,\ldots,S_m$. For each state $q \in Q$, define $L_q = \{w \in \Sigma^* : \delta(q_0,w) = q\}$. From the definition of the Myhill–Nerode relation it follows that each $L_q$ is a subset of some $S_i$.

The Myhill–Nerode theorem shows that there is a DFA with $m$ states, so suppose that ${\cal A}$ has $m$ states. From our comment above it follows that each $L_q$ equals some $S_i$. Rename the states so that $S_i = L_i$ (so now $Q = \{1,\ldots,m\}$). We now read off all the other data of ${\cal A}$:


The starting state $q_0$ is the one whose language contains the empty word $\epsilon$, that is, it is determined by the condition $\epsilon \in S_{q_0}$.
A state $i$ is accepting if $S_i \subseteq L$, and rejecting if $S_i \subseteq \overline{L}$ (it follows from the definition of the Myhill–Nerode relation that these are the only two possibilities).
Let $w_i$ be an arbitrary word in $S_i$, and let $\sigma \in \Sigma$. Then $\delta(i,\sigma) = j$ if $w_i\sigma \in S_j$. Again, the definition of the Myhill–Nerode relation shows that the choice of $w_i$ doesn't matter.


After renaming the states of ${\cal A}$ in a canonical way, we read off ${\cal A}$ from the Myhill–Nerode relation of $L$. It follows that all minimial DFAs for $L$ are equivalent.