Given $P(A) = 0.9$ and $P(B) = 0.9$, what is a lower bound of $P(A \cup B)$? if $P(A) = 0.9$ and $P(B) = 0.9$ what is the lower bound for $P(A \cup B)$?

My goal in 'this' answer is to supplement with some generalizations for those that are interested in such things.
A rule that is frequently taught in introductory probability is
$$P(A \cup B) = P(A) + P(B) - P(A\cap B)$$
where $A, B$ are two events from some event space $E$. This rule is accounting for potential overlap between events, and a natural way to generalize is to go from two events for a finite number of events. This is given by the inclusion-exclusion principle as applied in probability, which can be expressed as
$$P\left( \bigcup_{i=1}^{n} A_i\right) = \sum_{k=1}^{n} \left( (-1)^{k-1}\sum_{I\subseteq\{1,\cdots,n\}\ s.t\ |I|=k}  P(A_I)\right)$$
which expresses that the probability of the union as summing the probabilities of intersections of odd numbers of events and subtracting the probabilities of intersections of even numbers of events for all subsets of the powerset in cardinality up to $n$. This is one of those tools that can seem complicated at first, especially due to verbose notation, but follows a relatively simple pattern.
For calculating bounds, you can consider the Frechet inequalities:
$$\max \left( 0, \sum_{k=1}^n P(A_k) - (n-1) \right) \leq P\left( \bigcap_{k=1}^{n} A_k \right) \leq \min_k \{ P(A_k) \}$$
$$\max_k \{P(A_k)\} \leq P\left( \bigcup_{k=1}^{n} A_k \right) \leq \min \left(1, \sum_{k=1}^n P(A_k) \right)$$
In the two-event case, we can consider
$$\max \{P(A), P(B)\} \leq P\left( A \cup B \right)$$
which yields (from the OP)
$$\max \{0.9, 0.9\} = 0.9 \leq P\left( A \cup B \right)$$
which agrees with @BruceET's answer.
