The following estimate will be useful:
Claim. For any $a, b, x \geq 0$, we have
$$ \left| \frac{bx}{1+b^4x^2} - \frac{ax}{1+a^4x^2} \right| \leq \frac{3x|b-a|}{1+\min\{a,b\}^4x^2}. $$
Proof of Claim. Assume $a < b$ without losing the generality. Then by the fundamental theorem of calculus,
\begin{align*}
\left|\frac{bx}{1+b^4x^2} - \frac{ax}{1+a^4x^2}\right|
&\leq \int_{a}^{b} \left|\frac{\partial}{\partial \alpha} \left( \frac{\alpha x}{1+\alpha^4x^2} \right) \right| \, \mathrm{d}\alpha
= \int_{a}^{b} \left| \frac{x(1-3\alpha^4 x^2)}{(1+\alpha^4x^2)^2} \right| \, \mathrm{d}\alpha.
\end{align*}
Now using the triangle inequality, $\left|1-3\alpha^4 x^2\right| \leq 1+3\alpha^4x^2 \leq 3(1+\alpha^4x^2)$, and so, the last integral can be bounded from above by
\begin{align*}
&\leq \int_{a}^{b} \frac{3x}{1+\alpha^4x^2} \, \mathrm{d}\alpha
\leq (b-a) \max_{a \leq \alpha \leq b} \left(\frac{3x}{1+\alpha^4x^2}\right).
\end{align*}
Since the function $\alpha \mapsto 3x/(1+\alpha^4x^2)$ is decreasing in $\alpha$, the maximum is achieved at $\alpha = a$ and the desired inequality follows. ////
Returning to the original problem, note that
$$ \left| f_n(x) - f_{n+1}(x) \right| \leq \frac{3x}{1+n^4x^2} \leq \frac{3}{2n^2}, $$
where the first step is a consequence of the claim and the second step is simply an application of the AM-GM inequality. Therefore, by the Weierstrass M-test, the sum
$$ \sum_{k=1}^{n} (f_{2k-1}(x) - f_{2k}(x)) $$
converges uniformly on $[0, \infty)$, from which the uniform convergence of $S(x)$ easily follows.
Best Answer
Let $\varepsilon>0$ be given. Since $|f_k(x)|\leq g_k(x)$ for all $k$, we have
$$\Bigg|\sum_{k=n}^m f_k(x)\Bigg|\leq\sum_{k=n}^m|f_k(x)|\leq\sum_{k=n}^mg_k(x)$$
for all $m\geq n.$
By the Cauchy criterion for uniform convergence, a series $\sum_{k=1}^\infty f_k(x)$ converges uniformly on an interval $I$ if and only if for every $\varepsilon >0$ there exists an integer $N$ such that $\big|\sum_{k=n}^m f_k(x)\big|<\varepsilon$ whenever $x\in I$ and $m\geq n>N$.
Now, because $g_k(x)\geq0$ for all $k$ and $x$, this means that
$$\Bigg|\sum_{k=n}^m f_k(x)\Bigg|\leq\sum_{k=n}^mg_k(x)=\Bigg|\sum_{k=n}^mg_k(x)\Bigg|<\varepsilon$$ whenever $x\in I$ and $m\geq n>$ some integer $N$. Thus, $\sum_{k=1}^\infty f_k(x)$ is uniformly convergent on $[a, b].$