Given, $4(a-6)^2 + 9(b-5)^2 ≤ 36$. Let $a-6 = x$ and $b-5 = y$. Therefore, $4x^2 + 9y^2 ≤ 36$, which implies $\frac{x^2}{9} + \frac{y^2}{4} ≤ 1$. This is the equation of an ellipse.

According to set A, $|a-5| < 1$, which implies, since $a-6=x$, then $a-5 = x+1$. Thus $|x+1| < 1$, implying $-1 < x+1 < 1$, which leads to $-2 < x < 0$. Similarly, $|b-5| < 1$, implying $|y| < 1$, which leads to $-1 < y < 1$.

To check if the entire set A is inside of B, we can test a point of set A in the inequality $\frac{x^2}{9} + \frac{y^2}{4} ≤ 1$. Consider the point $(-2, 1)$. Substituting into the inequality gives LHS = $\frac{(-2)^2}{9} + \frac{1^2}{4} = \frac{4}{9} + \frac{1}{4} = \frac{25}{36} < 1$. Since the inequality holds, $(-2, 1)$ is inside the ellipse. Similarly, $(-2, -1)$ is also inside the ellipse. Hence, we can say that entire set A is inside of set B, therefore $A ⊂ B$.