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.
For the second case, take $f_n(x)=g_n(x) = x+\frac{1}{n}$. With $f(x)=g(x) = x$, we see that $f_n \to f, g_n \to g$ uniformly.
However, $f_n(x) g_n(x) = f(x)g(x) + \frac{2}{n} x + \frac{1}{n^2}$, hence the convergence is not uniform on unbounded sets.
Best Answer
What you are trying to prove is not true. There exist uniformly convergent sequences $f_n$ and $g_n$ such that their product sequence $f_n\cdot g_n$ converges uniformly to $f\cdot g$. Take $f_n$ and $g_n$ to be constant sequences, for example.
What your examples show is that the proposition in the linked question is false. The negation of that proposition is true, but it is different from the proposition you posted.