Congratulations! Your proof 2 is more direct, and it is correct.
If $\{f_n\}\subset L^+$, $f\in L^+$ such that $\{f_n\}$ is dominated by $f$ where $\int f < \infty$ then $$\limsup\int f_n\leq \int \limsup f_n$$
Proof 2 - Consider the sequence $\{f - f_n\}_n\subset L^+$ then applying Fatou's lemma we have
\begin{align*}
&\int (\liminf(f-f_n))\leq \liminf \int (f - f_n)\\
\Leftrightarrow &\int\liminf f - \int \liminf f_n \leq \liminf\int f - \liminf\int f_n\\
\Leftrightarrow &\int f - \int \limsup f_n \leq \int f - \limsup\int f_n
\end{align*}
Rearranging and cancelling out $\int f$ (which is possible since $\int f <\infty$) then we have $$\limsup\int f_n\leq \int \limsup f_n$$
Now let us look to your proof 1. The appproach you took here is viable, but there are some adjustments need.
Proof 1 - Consider the sequence $\{f - f_n\}\subset L^+$ then applying Fatou's lemma we have
$$\int \liminf(f-f_n)\leq \liminf \int (f - f_n)$$
We know that
$$\int\liminf (f - f_n) = \int \lim_{k\rightarrow \infty} \inf_{n\geq k}(f - f_n)$$
Since $\{\inf_{n\geq k}(f - f_n)\}_k$ is a non_decreasing sequence of non-negative functions in $L^+$ and $\{\inf_{n\geq k}(f - f_n)\}_k \to \lim_{k\rightarrow \infty} \inf_{n\geq k}(f - f_n)$, we have, from the Monotone Convergence theorem,
$$\int \lim_{k\rightarrow \infty}\inf_{n\geq k}(f-f_n) = \lim_{k\rightarrow \infty}\int \inf_{n\geq k}(f - f_n) \tag{1}$$
Now, since for all $k$ and $n\geq k$, $ \inf_{n\geq k}(f - f_n)\leq f_n$, we have that, for all $k$ and $n\geq k$,
$$ \int \inf_{n\geq k}(f - f_n)\leq \int (f-f_n)$$
So we have, for all $k$,
$$ \int \inf_{n\geq k}(f - f_n)\leq \inf_{n\geq k}\int(f- f_n) $$
So we have
$$ \lim_{k\rightarrow \infty}\int \inf_{n\geq k}(f - f_n)\leq \lim_{k\rightarrow \infty}\inf_{n\geq k}\int(f- f_n) \tag {2}$$
From $(1)$ and $(2)$, we have
$$\int \lim_{k\rightarrow \infty}\inf_{n\geq k}(f-f_n) \leq \lim_{k\rightarrow \infty}\inf_{n\geq k}\int(f- f_n) \tag{3}$$
Therefore we have,
\begin{align*}
\int f- \int\lim_{k\rightarrow \infty}\sup_{n\geq k} f_n&=\int (f- \lim_{k\rightarrow \infty}\sup_{n\geq k}f_n)=\int \lim_{k\rightarrow \infty}\inf_{n\geq k}(f-f_n)\leq \lim_{k\rightarrow \infty}\inf_{n\geq k} \int (f - f_n)= \\
&=\lim_{k\rightarrow \infty}\inf_{n\geq k} \left ( \int f - \int f_n \right)= \int f - \lim_{k\rightarrow \infty}\sup_{n\geq k}\int f_n\\
\end{align*}
Since $\int f<+\infty$, we can conclude that
$$\lim_{k\rightarrow \infty}\sup_{n\geq k}\int f_n \leq \int\lim_{k\rightarrow \infty}\sup_{n\geq k} f_n$$
that is to say:
$$\limsup \int f_n \leq \int\limsup f_n$$
Remark: Please note that your attempted proof 1, using Monotone convergence theorem is viable, but when we write down the details, we can see that we proving again Fatou's Lemma as part of proof 1. Please note that $(3)$ is essently Fatou's Lemma applied to $\{f-f_n\}_n$. And the rest of proof 1 after $(3)$ is essentially your proof 2. That is why Proof 2 is more concise, more direct and more elegant.
