Analyze statement I: A B A B.

$A \Delta B \subset A \cup B$.
The symmetric difference $A \Delta B$ is defined as the set of elements that are in A or B but not in their intersection. This can be written as $A \Delta B = (A-B) \cup (B-A)$.
The union $A \cup B$ contains all elements that are in A or B (or both).
If an element $x \in A \Delta B$, then $x \in (A-B)$ or $x \in (B-A)$.
If $x \in (A-B)$, then $x \in A$ and $x \notin B$. Since $x \in A$, it follows that $x \in A \cup B$.
If $x \in (B-A)$, then $x \in B$ and $x \notin A$. Since $x \in B$, it follows that $x \in A \cup B$.
In both cases, if $x \in A \Delta B$, then $x \in A \cup B$.
Therefore, $A \Delta B \subset A \cup B$ is true.

$A \cap B \subset A \Delta B$.
The intersection $A \cap B$ contains elements that are common to both A and B.
The symmetric difference $A \Delta B$ contains elements that are in A or B, but not in both. This means $A \Delta B = (A \cup B) - (A \cap B)$.
If an element $x \in A \cap B$, then $x$ is in both A and B. By the definition of symmetric difference, $x$ cannot be in $A \Delta B$.
For example, let $A = \{1, 2\}$ and $B = \{2, 3\}$.
Then $A \cap B = \{2\}$.
And $A \Delta B = (A-B) \cup (B-A) = \{1\} \cup \{3\} = \{1, 3\}$.
Clearly, $\{2\}$ is not a subset of $\{1, 3\}$.
Therefore, $A \cap B \subset A \Delta B$ is false.

$(A-B) \subset A$.
The set difference $A-B$ contains all elements that are in A but not in B.
By definition, every element in $A-B$ is an element of A.
Therefore, $(A-B) \subset A$ is true.

Statements I and III are true.

The option that states I and III are true is (D).

The final answer is $\text{D}$