The equivalence classes of an equivalence relation are nonempty, cover $A$, and are pairwise equal or disjoint; conversely every such cover arises from exactly one equivalence relation

For every $a\in A$ reflexivity gives $a\sim a$, that is $a\in [a]$; so each class is nonempty, each class is a subset of $A$ by construction, and each $a\in A$ lies in the class $[a]$, whence the union of the members of $A\mathrm{/}\sim$ is exactly $A$. This is claim 1.

Suppose $a\sim b$. If $c\in [b]$ then $b\sim c$, so $a\sim c$ by transitivity, so $c\in [a]$; thus $[b]\subseteq [a]$. Symmetry gives $b\sim a$, and the same argument with $a$ and $b$ interchanged gives $[a]\subseteq [b]$; hence $[a]=[b]$.

Every member of $P$ is nonempty, distinct members of $P$ are disjoint, and the union of $P$ is $A$; consequently each $a\in A$ lies in at least one $B\in P$, and in only one, since two members containing $a$ are not disjoint and therefore are not distinct. Write $B_a$ for that unique member.

Suppose $[a]=[b]$. By step 1.1, $b\in [b]=[a]$, which says $a\sim b$. Together with step 1.2 this is claim 2.

Suppose $[a]\cap [b]\mathrm{\ne }\mathrm{\varnothing }$ and pick $c$ in the intersection, so $a\sim c$ and $b\sim c$. Symmetry gives $c\sim b$, and transitivity then gives $a\sim b$, so $[a]=[b]$ by step 1.2. Hence for any $a,b$ either the classes are disjoint or they are equal, which is claim 3.

${\sim }_P$ is reflexive, since $a\in B_a$ gives $a{\sim }_{P}a$; it is symmetric, since the defining condition is symmetric in $a$ and $b$; and it is transitive, for if $a,b\in B$ and $b,c\in B^{\mathrm{\prime }}$ with $B,B^{\mathrm{\prime }}\in P$, then $b\in B\cap B^{\mathrm{\prime }}$, so $B$ and $B^{\mathrm{\prime }}$ are not disjoint and hence not distinct, giving $B=B^{\mathrm{\prime }}$ and $a,c\in B$. So ${\sim }_P$ is an equivalence relation on $A$.

Fix $a\in A$ and write $[a{]}_P$ for its class under ${\sim }_P$. If $b\in [a{]}_P$ then some $B\in P$ contains $a$ and $b$, and $B=B_a$ by the uniqueness in step 1.3, so $b\in B_a$; conversely if $b\in B_a$ then $B_a$ contains both $a$ and $b$, so $b\in [a{]}_P$. Hence $[a{]}_P=B_a$.

Every class of ${\sim }_P$ is a member of $P$ by step 3.1, and every $B\in P$ is nonempty, so choosing $a\in B$ gives $B=B_a=[a{]}_P$, a class; therefore the quotient set of ${\sim }_P$ is exactly $P$.

Uniqueness. Let $\approx$ be any equivalence relation on $A$ whose quotient set is $P$, and fix $a\in A$. Its class $[a{]}_{\approx }$ lies in $P$ and contains $a$ by step 1.1 applied to $\approx$, so $[a{]}_{\approx }=B_a$ by the uniqueness in step 1.3. Then $a\approx b$ holds exactly when $b\in [a{]}_{\approx }=B_a$, which by step 3.1 holds exactly when $a{\sim }_{P}b$; so $\approx$ and ${\sim }_P$ relate the same pairs and are equal as subsets of $A\times A$.

Claims 1, 2 and 3 are steps 1.1, 2.1 and 2.2; the converse is steps 2.3, 4.1 and 4.2.