The problem is that $-f_n$ increases to $-f$ which is not non-negative, so we can't apply directly to $-f_n$ the monotone convergence theorem. But if we take $g_n:=f_1-f_n$, then $\{g_n\}$ is an increasing sequence of non-negative measurable functions, which converges pointwise to $f_1-f$. Monotone convergence theorem yields:
$$\lim_{n\to +\infty}\int_X (f_1-f_n)d\mu=\int_X\lim_{n\to +\infty} (f_1-f_n)d\mu=\int_X f_1d\mu-\int_X fd\mu$$
so $\lim_{n\to +\infty}\int_X f_nd\mu=\int_X fd\mu$.
Note that the fact that there is an integrable function in the sequence is primordial, indeed, if you take $X$ the real line, $\mathcal M$ its Borel $\sigma$-algebra and $\mu$ the Lebesgue measure, and $f_n(x)=\begin{cases} 1&\mbox{ if }x\geq n\\\
0&\mbox{ otherwise}
\end{cases}$
the sequence $f_n $ decreases to $0$ but $\int_{\mathbb R}f_nd\lambda =+\infty$ for all $n$.
The monotone convergence theorem handles infinities gracefully, which can only be done for functions that are positive (or otherwise reasonably controlled from below). In particular, nowhere does it assume that $f_n$, $f$, or the integrals, are finite. This seems to be beyond the scope of Beppo Levi, so I'm not sure that fixing this issue alone is considerably easier than proving everything from scratch. But let me try.
Depending on your version of definitions, it may or may not be trivial that for a positive function $f$ the following special case of monotone convergence holds:
$$\intop_E f dm = \lim_{C \to +\infty, E_n \uparrow E} \intop_{E_n} \min(f, C) dm$$
where $E_n$ are sets of finite measure that approximate $E$ (I assume $\sigma$-finiteness; if it fails then we should restrict to $\{f > 0\}$; if it fails even there then monotone convergence holds almost trivially with both sides infinite).
Now in order to make use of Beppo Levi we should make the limit finite. I would do that by replacing $f_n$ by $f_{n,C,k} := (f_n \wedge C) \mathsf{1}[E_k]$ and $f$ by $f_{C,k} := (f \wedge C) \mathsf{1}[E_k]$ for some fixed $k$. Now we can safely apply Beppo Levi to the successive differences $f_{n+1,C,k} - f_{n,C,k}$ to obtain
$$\intop f_{C,k} dm = \lim_{n \to \infty} \intop f_{n,C,k} dm$$
Now take $C$ and $k$ to $\infty$ and interchange the limits. You can always do this with monotonely increasing limits (this is equivalent to rearrangement of terms in positive series, or Fubini on $\mathbb{N} \times \mathbb{N}$, or whatever you prefer). On the other hand, monotone convergence itself is about rearrangement or Fubini on $E \times \mathbb{N}$, so I'm not even sure you would view the things that I rely on as more basic than those that you prove...
Best Answer
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.)