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There's a nice application of the lemma to point-set topology (whether or not you consider this to be set theoretic is up to you :-) )

Define the following generalisation of separability:

A topological space X is ccc if every pairwise disjoint collection of non-empty open sets is countable.

It's clear that every separable space is ccc (because you can pick an element of the dense set in each open set), but ccc has the following theorem which makes it better behaved on products (a product of $> 2^{\aleph_0}$ spaces with more than one point is not separable):

Theorem: Let $\{ F_a : a \in A \}$ be a family of topological spaces such that every finite product is ccc. Then $\prod F_a$ is ccc.

Proof:

Let T be a an uncountable family of pairwise disjoint non-empty open sets. We can without loss of generality assume T consists of basis elements. each Each of these basis elements is a product of sets of the form $U_a \subseteq F_a$ with at most finitely many not equal to $F_a$.

Let $G = \{ \{ a : U_a \neq F_a \} : \prod U_a \in T \}$. This is an uncountable collection of finite sets, so contains a delta system, say with root R.

But we must have the projection of $T$ to $\prod_{a \in R} F_a$ still be disjoint: For any $a \not\in R$ we have $U_a \neq F_a$ for at most one element of T.

Thus But products of finitely many $T_a$ are ccc, thus we must have an uncountable collection of disjoint open sets in a finite product of the spacesprojection of T (and thus T itself) being countable, contradicting the hypothesis.

QED

This in particular gives us the following corollary:

Theorem: An arbitrary product of separables spaces is ccc.

Proof: A finite product of spearable spaces is separable and thus ccc.

which is interesting because e.g. it tells us that there are compact hausdorff spaces which are not the continuous image of $\{0, 1\}^\kappa$ for any $\kappa$.

1

There's a nice application of the lemma to point-set topology (whether or not you consider this to be set theoretic is up to you :-) )

Define the following generalisation of separability:

A topological space X is ccc if every pairwise disjoint collection of non-empty open sets is countable.

It's clear that every separable space is ccc (because you can pick an element of the dense set in each open set), but ccc has the following theorem which makes it better behaved on products (a product of $> 2^{\aleph_0}$ spaces with more than one point is not separable):

Theorem: Let $\{ F_a : a \in A \}$ be a family of topological spaces such that every finite product is ccc. Then $\prod F_a$ is ccc.

Proof:

Let T be a family of pairwise disjoint non-empty open sets. We can without loss of generality assume T consists of basis elements. each of these basis elements is a product of sets of the form $U_a \subseteq F_a$ with at most finitely many not equal to $F_a$.

Let $G = \{ \{ a : U_a \neq F_a \} : \prod U_a \in T \}$. This is an uncountable collection of finite sets, so contains a delta system, say with root R.

But we must have the projection of $T$ to $\prod_{a \in R} F_a$ still be disjoint: For any $a \not\in R$ we have $U_a \neq F_a$ for at most one element of T.

Thus we have an uncountable collection of disjoint open sets in a finite product of the spaces, contradicting the hypothesis.

QED

This in particular gives us the following corollary:

Theorem: An arbitrary product of separables spaces is ccc.

Proof: A finite product of spearable spaces is separable and thus ccc.

which is interesting because e.g. it tells us that there are compact hausdorff spaces which are not the continuous image of $\{0, 1\}^\kappa$ for any $\kappa$.