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 $A/{\sim}$ 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 $[a] \cap [b] \ne \varnothing$ 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 $b, c \in B'$ with $B, B' \in P$, then $b \in B \cap B'$, so $B$ and $B'$ are not disjoint and hence not distinct, giving $B = B'$ 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.