A counter-example for how the Downward Monotone Convergence theorem can fail is given by; Let $n = 1$ so we are working in $\mathbb{R}^1$ and let $E_n = [n,\infty)$. Then $m(E_n) = \infty$ for all $n$. And $E_n$ satisfies the criterion $E_1 \supset E_2 \supset E_3 \supset \cdots$. However, $m(\cap_{n=1}^{\infty} E_n) = m(\emptyset) = 0 \neq \infty = \lim \, m(E_n)$.
So let's go over the proof of the theorem to see exactly where the finiteness of an $E_n$ is used. First, using the same idea from the first problem we can write each $E_n$ as a disjoint union:
$$
E_n = \cup_{k=n}^{\infty} F_n \, \bigcup \, S
$$
Where each $F_n$ is given by $F_n = E_n - E_{n+1}$, and $S$ is given by $S = \cap^{\infty} E_n$. It should be clear that these sets are mutually disjoint. Now the additivity of measures gives
$$
m(E_n) = m(S) + \Sigma_{k=n}^{\infty} m(F_k)
$$
The statement that $m(E_n)$ is finite for some $n$ is exactly the statement that $\Sigma_{k=n}^{\infty} m(F_k)$ converges for some $n$ and that $m(S)$ is finite.
That $m(S)$ is finite is immediate from the monotonicity of measures and that $S \subset E_n$ for all $n$ along with that $E_n$ has finite measure for some $n$. And since the series $\Sigma_{k=n}^{\infty} m(F_k)$ converges for some $n$ we have:
$$
\lim_n \, \Sigma_{k=n}^{\infty} m(F_k) = 0.
$$
We get:
$$
\lim_n \, m(E_n) = m(S) + \lim_n \, \Sigma_{k=n}^{\infty} m(F_k) = m(S).
$$
Since $\mu_n \leqslant \mu$ for all $n$, we have
$$\int f\,d\mu_n \leqslant \int f\,d\mu$$
for all $n$, and therefore
$$\lim_{n\to\infty} \int f\,d\mu_n = \sup_n \int f\,d\mu_n \leqslant \int f\,d\mu.$$
Conversely, we have
$$\int f\,d\mu = \sup \Biggl\{ \int s\,d\mu : 0 \leqslant s \leqslant f,\; s\text{ simple}\Biggr\}.$$
So fix an arbitrary simple $s$ with $0 \leqslant s \leqslant f$. By the case for simple functions,
$$\int s\,d\mu = \lim_{n\to\infty} \int s\,d\mu_n \leqslant \lim_{n\to\infty} \int f\,d\mu_n.$$
Taking the supremum over all admissible $s$ gives
$$\int f\,d\mu \leqslant \lim_{n\to\infty} \int f\,d\mu_n,$$
so together with the first part, we have the monotone convergence theorem, supposing your argument for simple functions is correct. (It should be, that case is simple.)
Best Answer
The monotone convergence theorem holds for any measure space: wiki
The real numbers are (up to order isomorphism) the only ordered field with the least upper bound property (this property is called "Dedekind completeness") This is a well-known fact (wiki) and there is a proof in an appendix of Spivak's "Calculus".
Sneak answer: if you don't use in $k$ then the sets do not necessarily cover $\mathbb{R}$: let $g$ be the characteristic function of $[0,1]$. What if $f_n=(1-1/n)g$ and $\phi=g$? Also, the proof should not require completeness of the Lebesgue $\sigma$-algebra at any point.