Yes, easily. Consider all well-ordered sequences $(A_\alpha)$, of any length, such that each $A_\alpha$ is infinite and satisfying your almost containment condition. Make one such sequence less than another if the first is an initial segment of the second. Zornicate to get a maximal such sequence, and it is easy to show that a maximal sequence cannot have countable length.

(In response to YCor's comment, I assume you also meant to say that $A_\alpha\setminus A_{\alpha+1}$ is infinite for all $\alpha$.)

Yes, easily. Consider all well-ordered sequences $(A_\alpha)$, of any length, such that each $A_\alpha$ is infinite and satisfying your almost containment condition. Make one such sequence less than another if the first is an initial segment of the second. Zornicate to get a maximal such sequence, and it is easy to show that a maximal sequence cannot have countable length.

Yes, easily. Consider all well-ordered sequences $(A_\alpha)$, of any length, such that each $A_\alpha$ is infinite and satisfying your almost containment condition. Make one such sequence less than another if the first is an initial segment of the second. Zornicate to get a maximal such sequence, and it is easy to show that a maximal sequence cannot have countable length.

Yes, easily. Consider all well-ordered sequences $(A_\alpha)$, of any length, such that each $A_\alpha$ is infinite and satisfying your almost containment condition. Make one such sequence less than another if the first is an initial segment of the second. Zornicate to get a maximal such sequence, and it is easy to show that a maximal sequence cannot have countable length.

(In response to YCor's comment, I assume you also meant to say that $A_\alpha\setminus A_{\alpha+1}$ is infinite for all $\alpha$.)