A)
Let's say $\text{A} = \set{\set{2,3,4}} \text{ and B } = \set{\set{2,3}}$ then $\text{A -B} = \set{\set{2,3,4}} $ and moreover $\set{4} \not\subseteq \text{A-B}$.
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$\text{Let A} = \set{4, \set{2,3,4}} \text{ and B } = \set{\set{2,3}}$ then $\text{A -B} = \set{4, \set{2,3,4}} $ and $\set{4} \subseteq \text{A-B}$.
So I think option A is $\text{NOT always true and NOT always false}$.
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B) $\text{S} \subseteq \text{A} \cap\text{B} $ then $\text{S} \subseteq \text{A}$ $ \land \subseteq \text{B} $.
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Assume $\text{S} \subseteq \text{A} \cap \text{B} $.
<-->$\forall \text{x}( \text{x} \in \text{S} \mapsto \text{x} \in \text{A} \cap\text{B})$
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<--> $\forall \text{x}( \text{x} \in \text{S} \mapsto( \text{x} \in \text{A} \land \text{x} \in \text{B}))$
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$\leftrightarrow$ $\forall \text{x}( \text{x} \notin \text{S} \lor( \text{x} \in \text{A} \land \text{x} \in \text{B}))$
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$\leftrightarrow$ $\forall \text{x}( (\text{x} \notin \text{S} \lor \text{x} \in \text{A} )\land( \text{x} \notin S \lor \text{x} \in \text{B}))$
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$\leftrightarrow$ $\forall \text{x}( (\text{x} \in \text{S} \mapsto \text{x} \in \text{A} )\land( \text{x} \in S \mapsto \text{x} \in \text{B}))$
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$\leftrightarrow$ $\forall \text{x}( (\text{x} \in \text{S} \mapsto \text{x} \in \text{A} )) \land\forall \text{x}( \text{x} \in S \mapsto \text{x} \in \text{B})$
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$\leftrightarrow$ $\text{S} \subseteq \text{A} \land \text{S} \subseteq \text{B}$
So $\text{S} \subseteq \text{A} \cap \text{B} $ $\mapsto$ $\text{S} \subseteq \text{A} \land \text{S} \subseteq \text{B}$
So option B is $\text{TRUE.}$.
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C) Assume $\set{3,4} \subseteq \text{ A-B } $ and $ \set{1,2} \subseteq \text{B} $ and $\set{1,2,3,4} \not\subseteq A \cup B$.
$\mapsto \exists\text{x}( \text{x} \in \set{1,2,3,4} \land \text{x} \not\in A \cup B)$.
$\text{Such an $\text{x}$ can't be 1 or 2 because there are already in B so they also will be a part of}$ $A \cup B$ $\text{ and it can't be 3 or 4 because there are already in A so they also will be a part of}$ $A \cup B$.
So by proof by Contradiction
$\set{3,4} \subseteq \text{ A-B } $ and $ \set{1,2} \subseteq \text{B} $ $\mapsto$ $\set{1,2,3,4} \subseteq A \cup B$
So option C is true.
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D) Similarly one can give proof for option D, it is also a True statement.
So I think option A is closest one, rest of all are True.