How to prove the 2nd & 3rd conditions of outer measure? I have this question on outer measure from Richard Bass' book:

Prove that $\mu^*$ is an outer measure, given a measure space $(X, \mathcal A, \mu)$ and define
$$\mu^*(A) = \inf \{\mu(B) \mid A \subset B, B \in \mathcal A\}$$
for all subsets $A$ of $X$.

Here are what I have gone so far:
(1) The first condition is the easiest one: 
$$\begin{align}
\mu^*(\emptyset) &= \inf \{\mu(B) \mid \emptyset \subset B, B \in \mathcal A\}\\
&= \mu (\emptyset) \\
&= 0
\end{align}$$
(2) Now the second condition. Let $D, E \in X$ and $D \subset E$,
$$\begin{align}
\mu^*(D) &= \inf \{\mu(D') \mid D \subset D', D' \in \mathcal A\}\\
\mu^*(E) &= \inf \{\mu(E') \mid E \subset E', E' \in \mathcal A\}\\
\end{align}$$
Here I need to prove $\mu^* (D) \leq \mu^*(E)$. It looks to me so intuitive especially if I draw Venn diagrams of $D, E, D'$ and $E'$, but I don't know how to say it in math-speak. I would appreciate helps on this 2nd. condition. 
(3) And this 3rd. condition is my major stumbling block: Given $(A_i)_{i \in \mathbb N} \subset X$, I need to arrive at
$$\mu^* (\bigcup _{i=1}^{\infty} A_i)\leq \sum_{i=1}^{\infty} \mu^* (A_i).$$
Here, I know for sure I need to  state this first: $\forall A_i, \exists B_i $ such that $ A_i \subset B_i, B_i \in \mathcal A$, but I don't think the next step is right:
$$\begin{align}
\mu^*(\bigcup_{i=1}^{\infty}A_i) &= \inf \{\bigcup_{i=1}^{\infty}\mu(B_i) \mid A_i \subset B_i, B_i \in \mathcal A\}\\
&= \ldots\\
\end{align}$$
I would appreciate any help on this 3rd. condition in addition to the 2nd. above. Thank you for your time and effort.

POST SCRIPT: After I posted this question, I found this proposition on the same text, perhaps this proposition holds key to the solution, in that I don't have to prove the 2nd and 3rd conditions? Thanks again.

Proposition: Suppose $\mathcal C$ is a collection of subsets of $X$ such that $\emptyset$ and $X$ are both in $\mathcal C$. Suppose $\mathscr l : \mathcal C \to [0, \infty]$ with $\mathscr l (\emptyset) = 0$. Define
$$\mu^* (E) = \inf \{ \sum_{i=1}^{\infty} \mathscr l (A_i) \mid A_i \in \mathcal C \text{ for each } i, \text{and }  E \subset \cup_{i=1}^{\infty} A_i \}.$$
  Then $\mu^*$ is an outer measure.

 A: I do not see how the quoted proposition helps, although I may be missing something. Nonetheless, the exercise is easy to solve without it.
(2) Since $D\subset E$, $X_E:=\{E'\in\mathcal A\ :\ E\subset E'\}\subset X_D:=\{D'\in\mathcal A\ :\ D\subset D'\}$, so the infimum over the elements of $X_E$ is larger than the infimum over $X_D$. In other words,
$$
\mu^*(E)=\inf_{B\in X_E}\{\mu(B)\}\ge\inf_{B\in X_D}\{\mu(B)\}=\mu^*(D).
$$
(3) Let $(A_i)_{i\in\mathbb N}\subset X$. By definition of $\mu^*$, for each $\varepsilon>0$ and $n\in\mathbb N$, there exists $B_n\in\mathcal A$, $B_n\supset A_n$, such that
$$
\mu(B_n)<\mu^*(A_n)+\frac\varepsilon{2^n}.
$$
Then $\bigcup_nB_n\in\mathcal A$ and $\bigcup_nB_n\supset\bigcup_nA_n$, so again by definition of $\mu^*$,
$$
\mu^*\left(\bigcup_nA_n\right)
\le\mu\left(\bigcup_nB_n\right)
\le\sum_{n\in\mathbb N}\mu\left(B_n\right)
<\sum_{n\in\mathbb N}\mu^*\left(A_n\right)+\varepsilon,
$$
and you conclude since $\varepsilon$ is arbitrary.
A: At the same time I received response from @Ian, I came up with my own solution, I am posting it here to let others see if its logic makes sense. If it does, then I would value @Ian's for its elegance and brevity, and mine for its originality (but folksy and clumsy.) Thanks again to @Ian for responding.

Here is my solution after some changes per @Ian's correction (see comment below): 
(1) The firs condition:
$$\begin{align}
\mu^*(\emptyset) &= \inf \{\mu(B) \mid \emptyset \subset B, B \in \mathcal A\}\\
&= \mu (\emptyset) \\
&= 0
\end{align}$$
(2) The second condition: Let $D, E \subset X$ and $D \subset E$, then
$$\begin{align}
\mu^* (D) &= \inf \{ \mu (D') \mid D \subset D', D' \in \mathcal A\}, 
\\
\mu^* (E) &= \inf \{ \mu (E') \mid E \subset E', E' \in \mathcal A\}.
\end{align}$$
From $D \subset E$ we need to show that 
$$\inf\{\mu (D') \mid D' \in \mathcal A, D \subset D'\} \leq \inf \{\mu(E') \mid E'\in \mathcal A, E \subset E'\}.$$
Suppose that by way of contraidiction, $D \subset E$ leads to $\inf \{\mu (D')\} > \inf \{\mu (E')\}$ instead. But this contradicts the definition of $D′$ being the infimum of all sets belong to $\mathcal A$ and containing $D$. Therefore $D \subset E$ has to lead to $\inf\{\mu (D′)\}\leq \inf \{\mu (E′)\}$.
Hence $\mu^* (D) \leq \mu^* (E)$.
(3) The third condition: Let $(A_i)_{i \in \mathbb N} \subset X$, and  $(B_i)_{i \in \mathbb N} \in \mathcal A.$
$$\begin{align}
\text{Given } \mu^* (A_i) &= \inf \{\mu(B_i) \mid A_i \subset B_i, B_i \in \mathcal A\}, \\
\text{therefore, } \sum_{i=1}^{\infty}\mu^* (A_i) &= \sum_{i=1}^{\infty}\inf \{\mu(B_i) \mid A_i \subset B_i, B_i \in \mathcal A\}.\\
\text{On the other hand, } \ \mu^* (\bigcup_{i=1}^{\infty}A_i) &= \inf \{\mu(\bigcup_{i=1}^{\infty} B_i) \mid \bigcup_{i=1}^{\infty} A_i \subset \bigcup_{i=1}^{\infty} B_i, B_i \in \mathcal A\}.\\
\end{align}$$
CASE-1: If $B_i$'s are pairwise disjoint, that is $B_i \cap B_j = \emptyset$ for all $i, j \in \mathbb N$, then
$$\sum_{i=1}^{\infty} \mu^* (A_i) = \mu^* (\bigcup_{i=1}^{\infty} A_i).$$
CASE-2: If $B_i \cap B_j \neq \emptyset$ for certain $i, j \in \mathbb N$, then
$$\sum_{i=1}^{\infty} \mu^* (A_i) < \mu^* (\bigcup_{i=1}^{\infty} A_i).$$
Therefore for all cases
$$\sum_{i=1}^{\infty} \mu^* (A_i) \leq \mu^* (\bigcup_{i=1}^{\infty} A_i),$$
hence $\mu^*$ is an outer measure. $\qquad \blacksquare$
