For any measurable set $E \subset \mathbb{R},$ we have $\displaystyle \int_Ef \leqslant \int_{\mathbb{R}}f < \infty$ and $\displaystyle \int_{E}f_n \leqslant \int_{\mathbb{R}}f_n \to \int_{\mathbb{R}}f$.
Hence, for sufficiently large $n$, we have
$$\displaystyle \int_Ef_n < \infty.$$
Using Fatou's Lemma,
$$\int_E f = \int_E \lim f_n=\int_E \liminf f_n \leqslant \liminf \int_E f_n$$
and reverse Fatou's Lemma,
$$\liminf \int_E f_n \leqslant \limsup\int_E f_n \leqslant \int_E \limsup f_n = \int_E \lim f_n = \int_Ef.$$
Hence,
$$\tag{*}\int_Ef \leqslant \liminf \int_E f_n \leqslant \limsup\int_E f_n \leqslant\int_Ef,$$
and
$$\liminf \int_E f_n = \limsup\int_E f_n=\lim \int_E f_n = \int_Ef.$$
Update:
We can avoid the argument based on reverse Fatou as follows.
Note that since $f_n \to f$ and $\int_{\mathbb{R}} f_n \to \int_{\mathbb{R}} f$ we have
$$\begin{align}\limsup \int_E f_n &= -\liminf \left(-\int_E f_n \right) \\&= -\liminf \left(\int_{\mathbb{R}\setminus E} f_n-\int_{\mathbb{R}} f_n \right) \\ &= -\liminf \int _{\mathbb{R}\setminus E} f_n+\liminf \int_{\mathbb{R}} f_n \\ &\leqslant -\int_{\mathbb{R} \setminus E} \liminf f_n + \int_{\mathbb{R}} f \\ &= -\int_{\mathbb{R} \setminus E} f + \int_{\mathbb{R}} f \\ &= \int_E f\end{align} $$
The chain of inequalities (*) now follows.
$\int (f-f_n)^{+} \to 0$ by DCT because $(f-f_n)^{+} \leq f$ and $(f-f_n)^{+} \to 0$. Also $\int (f-f_n) \to 0$ by hypothesis. Subtract the first from the second to get $\int (f-f_n)^{-} \to 0$. Add this to $\int (f-f_n)^{+} \to 0$ to get $\int |f-f_n| \to 0$. For any measurable set $E$ we have $\int_E |f-f_n|\leq \int_{\mathbb R} |f-f_n| \to 0$ which implies $\int_E f_n \to \int_E f$.
Best Answer
The proof you mention seems wrong to me, since it uses the inequality $$ \limsup\int_Ef_n\le\int_E\limsup f_n $$ You may complete the proof applying Fatou's lemma to $E^c=\mathbb{R}\setminus E$. $$ \begin{multline} \int_Ef=\int_{\mathbb R}f-\int_{E^c}f\ge\int_{\mathbb R}f-\liminf\int_{E^c}f_n\\=\int_{\mathbb R}f-\liminf\Bigl(\int_{\mathbb R}f_n-\int_{E}f_n\Bigr)=\limsup\int_Ef. \end{multline}$$