Q:

Find $\bigcup_{n=1}^\infty A_n$ when $A_n = \big(a,b - \frac{b-a}{2n}\big]$

The solution is fairly obvious but I have a hard time convincing myself that I can justify each step instead of just knowing the answer. Here's what I have:

Note that $\forall n\in\Bbb N$, we have $A_{n} \cup A_{n+1} = A_{n+1}$
Therefore, $\bigcup_{n=1}^\infty A_n = \lim \limits_{n \to \infty} A_1 \cup A_2 \cup \dots \cup A_n = \lim \limits_{n \to \infty}A_n$
Using the definition of $A_n$, $\lim \limits_{n \to \infty} (a,b-\frac{b-a}{2n}] = (a , b - \lim \limits_{\epsilon \to 0}\epsilon] = (a,b)$

For some context, I am in the first week of a Statistics masters and have no formal math background. Therefore, I am not completely sure if my proposed steps are omitting/assuming any crucial information or perhaps misusing notation.
My specific gripes comes with steps 1 and 3:
In the first case, would I be expected to prove that $A_n \subset A_{n+1}$? I could provide a specific breakdown of that logic statements defining the lhs and rhs statements to show their equivalence, but this seems overkill.
For the last step, I am completely stumped regarding how to handle the limit and interval boundaries. My calculus I training tells me that because $\lim \limits_{\epsilon \to 0}\epsilon = 0$, the interval should evaluate to $(a,b]$ but I know that logically that is wrong. However, I can't come up with a statement to justify the difference.

A:

We can try the following approach (which was also suggested by @quasi).
First, let us observe that for each $n$, $$\frac{b-a}{2n}>0$$ and so 
$$b-\frac{b-a}{2n}<b$$
implying that  $A_n\subset (a,b)$ and hence
$$\bigcup_{n=1}^\infty A_n\subset (a,b).\tag 1$$
Next, let $x\in (a,b)$. Using the Archimedean Property, we can find $m\in\Bbb N$ such that $$\frac{1}{m}<\frac{2(b-x)}{b-a}.$$
This means that 
$$a<x<b-\frac{b-a}{2m}$$ for some $m\in\Bbb N$. Hence,
$$x\in\bigg(a,b-\frac{b-a}{2m}\bigg)=A_m.$$
But 
$$A_m\subset \bigcup_{n=1}^\infty A_n.$$
This proves that 
$$(a,b)\subset \bigcup_{n=1}^\infty A_n.\tag 2$$
Combining $(1)$ and $(2)$, we get
$$\bigcup_{n=1}^\infty A_n=(a,b).$$
NOTE: It should be noted that $b\notin A_n$ for all $n\in\Bbb N$.

