How would I prove the following containment relation about the interior of a union:
(A $\displaystyle \cup$ B)º $\displaystyle \supseteq$ Aº $\displaystyle \cup$ Bº
Here's one way to go about it. Pick an arbitrary $\displaystyle x\in\text{int}(A)\cup\text{int}(B).$ Then, WLOG, $\displaystyle x\in\text{int}(A).$ This implies that there is an $\displaystyle \varepsilon>0$ such that an open ball centered at $\displaystyle x$ of radius $\displaystyle \varepsilon$ is contained in $\displaystyle \text{int}(A).$ Now show that both $\displaystyle x\in A\cup B,$ and the ball is also contained in $\displaystyle A\cup B.$ That should do it, right?
You can call it a neighborhood, or a sphere, or whatever you want. On the real line, it'd be an open interval of length $\displaystyle 2\varepsilon$ centered at $\displaystyle x.$ In 2 dimensions, it'd be a circle of radius $\displaystyle \varepsilon$ centered at $\displaystyle x$ (using the usual Euclidean length). In 3D, it'd be a sphere of radius $\displaystyle \varepsilon$ centered at the vector $\displaystyle x.$ And so on. Does that make sense? It's a small open set containing $\displaystyle x.$ That's the critical part.
If you're doing real analysis on the real line, you could write your neighborhood like this: $\displaystyle (x-\varepsilon,x+\varepsilon).$ You want to show that
$\displaystyle x\in A\cup B,$ and that
$\displaystyle (x-\varepsilon,x+\varepsilon)\subseteq A\cup B.$
Right?