Introduction. We first need to recall a few facts from basic field theory. For the definition of separability see the introduction section in this post. We have the following facts.

Fact 1. Let $E/F$ be an algebraic field extension and $a \in E.$ Then $F(a)/F$ is separable if and only if $a$ is separable over $F.$

Fact 2. Let $F \subseteq E \subseteq L$ be a chain of fields and suppose that $L/F$ is algebraic. If both $E/F$ and $L/E$ are separable, then $L/F$ is separable.

Fact 3. If $E/F$ is a separable field extension and $[E:F] < \infty,$ then $E=F(a)$ for some $a \in E.$

Theorem. Let $D$ be a division algebra with the center $k$ and suppose that $\dim_k D < \infty.$ There exists $a \in D$ such that $K=k(a)$ is a maximal subfield of $D$ and $K/k$ is separable.

Proof. Let $A$ be the set of all subfields of $D$ which are separable extensions of $k.$ This set is non-empty because $k \in A.$ Since $\dim_k D < \infty,$ the set $A$ with $\subseteq$ has a maximal element, say $K.$ Let $C(K)$ be the centralizer of $K$ in $D.$ Suppose that $C(K) \neq K.$ Then $C(K)$ is a noncommutative division ring and $K \subset C(K).$ Let $Z(C(K))$ be the center of $K.$ Then, since $K$ is commutative, we have $Z(C(K))=C(C(K))=K,$ by the double centralizer theorem. Clearly $C(K)$ is algebraic over $K$ because $[C(K):K] < \infty.$ Hence, by the Jacobson-Noether theorem, there exists $a \in C(K) \setminus K$ such that $a$ is separable over $K.$ So we have the chain of fields $k \subset K \subset K(a)$ where both $K/k$ and $K(a)/K$ are separable (see Fact 1). Thus, by Fact 2, $K(a)/k$ is separable and so $K(a) \in A.$ But this contradicts the maximality of $K$ in $A.$ This contradiction implies $C(K)=K$ and so, by Corollary 3, $K$ is a maximal subfield of $D.$ Finally, by Fact 3, $K=k(a)$ for some $a \in K. \ \Box$