Does excision imply that these inclusions are isomorphisms in homology? In exercise $4$, page 230 of Bredon, he asks for a proof of the Mayer-Vietoris sequence using a commutative braid diagram which substitutes some terms by others using excision. I've solved the exercise assuming the validity of the diagram, but it is not clear to me why the excision step is true.
The context is the following. We have $X=\text{int}(A)\cup \text{int}(B)$ (no assumption is made over $A,B$). He then proceeds to say that the inclusions
$$(A,A \cap B) \hookrightarrow (A \cup B, B)$$
and
$$(B, A \cap B) \hookrightarrow (A \cup B, A)$$
provide isomorphisms via excision. But the excision axiom  states that we must have the following situation:

$U \subset K$, $U$ open in $X$ such that $\overline{U} \subset \text{int}(K).$

But when we excise in the previous situation, we are eliminating (for instance, in the first example) the set $B-A$, which is not even necessarily open. 
Therefore, my question is: Can we really use excision, as stated, in this situation? If so, how?
EDIT: As pointed out in the comments, excision holds in less restrictive situations in singular homology. But the point of this exercise is to prove that "Therefore, the Mayer-Vietoris sequence follows from the axioms alone" (in Bredon's words). And in the axioms for homology, he states clearly:

"Given the pair $(X,A)$ and an open set $ U \subset X$ such that $\overline{U} \subset \text{int}(A)$ (...)"

 A: The excision axiom in its general form (each $U$ such that $\overline{U} \subset int(A)$ can be excised) is equivalent to the following statement:
If $X=\text{int}(A)\cup \text{int}(B)$, then the inclusion
$$j : (A,A \cap B) \hookrightarrow (A \cup B, B) = (X,B)$$
induces isomorphisms in homology.
This is an easy excercise. See also for example Proposition 7.2 in
Switzer, Robert M. Algebraic topology-homotopy and homology. Springer, 2017.
This shows that under the assumption in Bredon's exercise (excision axiom for open sets) you cannot prove that $j$ induces isomorphisms in homology.
Added:
As Cave Johnson pointed out in his comment, my above claim that you cannot prove that $j$ induces isomorphisms in homology is somewhat courageous. In fact, this claim means that the excision axiom for open sets is strictly weaker than the general excision axiom. 
Let us look to the classic
Eilenberg, Samuel, and Norman Steenrod. Foundations of algebraic topology
from 1952. In Chapter I the axioms of a homology theory are introduced. The excison axiom (Axiom 6) occurs in the "open" variant. It is shown that it is equivalent to  
Axiom 6'. Let $X_1$ and $X_2$ be subsets of a space $X$ such that $X_1$ is
closed and $X = int(X_1) \cup int(X_2)$ . If $i: (X_1,X_1 \cap X_2) \to (X_1  \cup X_2,X_2)$ is admissible, then it induces isomorphisms $i_\ast : H_q (X_1,X_1 \cap X_2) \approx H_q(X_l \cup X_2,X_2)$ for each $q$.
Here, being admissible means that both pairs of spaces belong to the admissible category $\mathfrak{A}$ on which the homology theory is defined (Eilenberg and Steenrod do not restrict to the category $\mathfrak{T}^2$ of all pairs of topological spaces).
In Chapter VII Theorem 9.1 they show that singular homology theory satisfies the "general" excision axiom for arbitrary $U$ and close with the following remark:
It should be pointed out that the excision axiom that has been
obtained for the singular theories is stronger than that required in I,3,
namely, the assumption that $U$ be an open subset of $X$ was not made.
A discussion about various variants of excision can be found in Chapter X Section 5. The variants considered by us occur as (E) and (E$_2$).
I do not know whether they are equivalent on $\mathfrak{T}^2$, but I doubt it.
My observation is that in the literature one most frequently finds the excision axiom in its stronger form, but it seems to depend on the taste of the author. For practical applications it doesn't make much difference anyway. 
