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Is there a set $A\ \subseteq P(\{x_1, x_2, x_3, \ldots \}) $ of cardinality $\aleph$ such that for every two sets $B,C\in A$: $B\ \cap C=B $ or $B\ \cap C=C $ ?

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Let $A=\{\{x_1\},\{x_1, x_2\}, \{x_1, x_2, x_3\}, \ldots \} $

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You should look into ordinals. An ordinal is a hereditarily transitive set - that is, $\alpha$ is an ordinal if

  • Whenever $x\in y\in\alpha$, we also have $x\in\alpha$ ($\alpha$ is transitive); and

  • Whenever $z\in\alpha$, and $x\in y\in z$, then $x\in z$ (ever element of $\alpha$ is transitive).

(There are other equivalent definitions, but I find this one the easiest to think about.) Some examples of ordinals are:

  • $\emptyset$ is (trivially) an ordinal.

  • $\{\emptyset\}$ is also an ordinal.

  • If $\alpha$ is an ordinal, then so is $\alpha\cup\{\alpha\}$.

  • And if $\alpha_i$ ($i\in I$) are all ordinals, then their union is also an ordinal.

It's a good exercise to show that every element of an ordinal is an ordinal. Additionally, it turns out that the ordinals are linearly ordered (in fact, well-ordered) by $\in$; these two facts imply that given any two ordinals $\alpha,\beta$, either $\alpha\subseteq\beta$ or $\beta\subseteq \alpha$. So since there are arbitrarily large ordinals, this gives you the result you're looking for, regardless of what cardinality you're looking for.

Another good example is provided by linear orders in general - like $\mathbb{Q}$ or $\mathbb{R}$.

Let's look at $\mathbb{Q}$ to be concrete. For $q\in \mathbb{Q}$, let $$A_q=\{p\in\mathbb{Q}: p

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    Actually the last example also works with $q\in\Bbb R$, i.e., we have an uncountable family of countable sets with the desired property. – 2017-02-12
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    @HagenvonEitzen Indeed, I didn't mean to imply that it only worked for $\mathbb{Q}$. – 2017-02-12