= Solution
For the standard definition of <algebraic independence>, finite tuples are required to have distinct entries. Let $\mathcal P$ be the set of all subsets of $L$ with <algebraic independence> over $K$, ordered by inclusion. It is nonempty because $\varnothing\in\mathcal P$.
If $\mathcal C$ is a chain in $\mathcal P$, put $U=\bigcup\mathcal C$. A finite tuple of distinct elements of $U$ lies in one member of $\mathcal C$: choose a member containing each element and take the largest of those finitely many members. A polynomial relation on that tuple would then be a polynomial relation in that member, contradicting its <algebraic independence>. Thus $U\in\mathcal P$ and is an upper bound. For the empty chain, $\varnothing$ is an upper bound. The <Zorn's lemma> now gives a maximal member $B$, which is a <transcendence basis> by the specified maximality criterion. To also recover the usual field-theoretic definition, if $a\in L$ were transcendental over $K(B)$, clearing denominators in a polynomial relation would show that $B\cup\{a\}$ still has <algebraic independence> over $K$, contradicting maximality. Thus $L$ is algebraic over $K(B)$. Hence \b[every field extension has a <transcendence basis>].
There is a literal issue in the printed definition: it does not require the tuple entries to be distinct. With that wording, any nonempty $X$ fails the test: for $a\in X$, the nonzero polynomial $X_1-X_2$ vanishes on $(a,a)$, in every characteristic. Thus under the literal printed test the family is exactly $\{\varnothing\}$, and $\varnothing$ is its maximal member. The same chain-and-<Zorn's lemma> argument still proves existence for that literal notion, but it is not the usual <transcendence basis> notion. \b[The conventional result above uses distinct tuple entries.]
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