Let $A$ and $B$ be integral domains with $A\subset B$. The transcendence degree of $B$ over $A$ is defined as the transcendence degree of $Quot(B)$ over $Quot(A)$. Denote it by $trdeg(B/A)$. Let $S$ be a subset of $B$ that is algebraically independent over $A$. Then $S$ is algebraically independent over $Quot(A)$. Thus $Card(S) \leq trdeg(B/A)$.
Is it true that there is a maximal algebraically independent subset of $B$ over $A$? If $S$ would be a maximal algebraically independent subset of $B$, would it then follow that $Card(S) = trdeg(B/A)$? In the case where $trdeg(B/A)$ is finite, $Card(S)$ attains a maximum value, but does this imply that $Card(S) = trdeg(B/A)$?
The fact that there is a maximal algebraically independent subset of $B$ over $A$ can be proved using Zorn's lemma in the same way as if $A$ and $B$ were fields.
If $S$ is a maximal algebraically independent subset of $B$ over $A$, then every element of $B$ is algebraic over $A[S]$. Therefore every element of $\operatorname{Quot}(B)$ is algebraic over $A[S]$, hence over $\operatorname{Quot}(A[S]) = \operatorname{Quot}(A)(S)$. Thus $S$ is a maximal algebraically independent set in $\operatorname{Quot}(B)$ over $\operatorname{Quot}(A)$. By definition then, $\operatorname{Card}(S)$ is the transcendence degree of $B$ over $A$.
In your last part, the answer is yes, by definition of transcendence degree and the foregoing arguments.