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...
Your sequence converges uniformly on each $F_m^c$. Hence it converges pointwise on each $F_m^c$. Hence it converges pointwise on the union of the $F_m^c$. (This is a basic special property of pointwise convergence: if $f_n$ converges pointwise on each set in a family, then it converges pointwise on their union.) But the union of the $F_m^c$ has full measure.
In terms of a $\varepsilon,n_0$ proof, let $x \in F^c$, then choose $m$ such that $x \in F_m^c$. Since you have uniform convergence on $F_m^c$, given $\varepsilon > 0$ you get $n_0$ so that if $x \in F_m^c$ and $n \geq n_0$ then $|f_n(x)-f(x)|<\varepsilon$.
Best Answer
The given example $f_n = 1_{[1,\frac 1n]} f$ , $f = \frac 1x$ on $(0,1]$ serves as a counterexample.
The reason is that if the $f_n$ are uniformly integrable, then their integrals don't blow up , because you can split your domain into two parts by cutting off $f_n$, one which is controlled by uniform integrability (a "large $f_n$" part) and the other where you control the integral of the $f_n$ because $f_n$ is "small" there. Pushing such an inequality(which holds for all $n$ because of u.i) to the limit $n \to \infty$ gives that any pointwise limit would be integrable. Without uniform integrability, you would not be able to write the inequality at all.