Here's your Venn diagram:
All of the colored regions (red and blue), taken together, represent the event $A \cup B.$
The event $A \cap B$ is represented by the red region alone.
This shows that $A \cap B \subseteq A \cup B$; everything in the intersection of the sets is also in their union.
You can't be in $A\cap B$ without also being in $A \cup B.$
It follows that $P(A\cap B) \leq P(A \cup B).$ This is always true.
Now what can we conclude if it is also true that
$P(A\cup B) \leq P(A \cap B)$?
If both inequalities are true, we can put them together like this:
$$P(A\cap B) \leq P(A \cup B)\leq P(A \cap B).$$
Notice how we have $P(A \cap B)$ on both ends. That is, we have a chain of inequalities like this: $x \leq y \leq x.$
The only way for this to be true is if $x = y.$
Conclusion: $P(A\cap B) = P(A \cup B).$
The probability of anything occurring in the blue regions is zero.
The only way for any of the events $A$, $B$, or $A\cup B$ to occur is when $A\cap B$ occurs; and if $A\cap B$ occurs then the other three events have also occurred.
Since all four events either occur together or none of them occurs,
we have $P(A\cap B) = P(A \cup B) = P(A) = P(B).$