The solution depends on the definition of uniform integrability you use. The most convenient one states that:
A collection $H$ of integrable functions bounded in $L^1(\Omega,\mathcal F,\mu)$ is uniformly integrable if $\int\limits_A|h|\to0$ when $\mu(A)\to0$, uniformly on $h$ in $H$, that is,
$$
\forall\varepsilon\gt0,\ \exists\eta\gt0,\ \forall A\in\mathcal F,\ \forall h\in H,\ \mu(A)\leqslant\eta\implies\int_A|h|\mathrm d\mu\leqslant\varepsilon.
$$
Then the proof is direct. First, if $\alpha H=\{\alpha h\mid h\in H\}$ and $H$ is uniformly integrable, then $\alpha H$ is bounded in $L^1$ and, if $\eta(\varepsilon,H)$ makes the implication above true for $H$ and $\varepsilon$, for each $\varepsilon$, then $\eta\left(\frac{\varepsilon}{|\alpha|},H\right)$ makes it true for $\alpha H$ and $\varepsilon$.
Second, if $H+K=\{h+k\mid h\in H,\ k\in K\}$ and $H$ and $K$ are uniformly integrable, then $H+K$ is bounded in $L^1$ and $\min\{\eta(\varepsilon,H),\eta(\varepsilon,K)\}$ makes the implication above true for $H+K$ and $2\varepsilon$.
Hence for every uniformly integrable collections $H$ and $K$ and every scalar $\alpha$ and $\beta$, $\alpha H+\beta K$ is uniformly integrable as well.
Note finally that the sequences $(f_n)_{n\in\mathbb N}$ and $(g_n)_{n\in\mathbb N}$ are uniformly integrable if and only if the collections $H=\{f_n\mid n\in\mathbb N\}$ and $K=\{g_n\mid n\in\mathbb N\}$ are uniformly integrable and that, then, $\alpha H+\beta K$ contains every function $\alpha f_n+\beta g_n$ (and many more). Hence the uniform integrability of $\alpha H+\beta K$ implies the one of $\{\alpha f_n+\beta g_n\mid n\in\mathbb N\}$.
If one wishes to use the definition that:
A collection $H$ is uniformly integrable if and only if $\int\limits_{|h|\gt t}|h|\mathrm d\mu\to0$ when $t\to+\infty$, uniformly on $h$ in $H$, that is,
$$
\forall\varepsilon\gt0,\ \exists t,\ \forall s,\ \forall h\in H,\ s\geqslant t\implies\int_{|h|\geqslant s}|h|\mathrm d\mu\leqslant\varepsilon,
$$
then the proof is similar but perhaps less convenient. Denote by $t(\varepsilon,H)$ any value of $t$ making the implication above true for $H$ and $\varepsilon$. Then $t(\varepsilon,H)$ works for $\alpha H$ and $|\alpha|\varepsilon$ hence $H$ uniformly integrable implies $\alpha H$ uniformly integrable. For the sum $H+K$, one can convince oneself of the validity of the pointwise inequality
$$
|h+k|\,\mathbf 1_{|h+k|\geqslant 2t}\leqslant 2|h|\,\mathbf 1_{|h|\geqslant t}+2|k|\,\mathbf 1_{|k|\geqslant t},
$$
and deduce that, in our notations,
$\max\{t(\varepsilon,H),t(\varepsilon,K)\}$ works for $H+K$ and $4\varepsilon$. QED.
Best Answer
You misunderstood what he says. Folland writes:
So $f_n=\chi_{(n,n+1)}$ does not converge uniformly, but pointwise which is evident. Only the first one converges uniformly.