Unlike $S_n(x) = \sum_{k=1}^n \sin kx$, the sum $\sum_{k=1}^n \frac{\sin kx}{k}$ is uniformly bounded for all $n$ and all $x \in \mathbb{R}$.
Hence, this series converges uniformly for all $x \in \mathbb{R}$ by the Dirichlet test -- since $(\ln k)^{-1} $ converges to $0$ monotonically and uniformly with respect to $x$.
Proving that $\sum_{k=1}^n \frac{\sin kx}{k}$ is uniformly bounded requires some effort. Because of periodicity, we can consider WLOG $x \in (0,\pi)$.
With $m = \lfloor1/x \rfloor$ we have
$$\left|\sum_{k=1}^n \frac{\sin kx}{k}\right| \leqslant \sum_{k=1}^{m}\frac{|\sin kx|}{k} + \left|\sum_{k=m+1}^{n}\frac{\sin kx}{k}\right| $$
For the first sum on the RHS,
$$\sum_{k=1}^{m}\frac{|\sin kx|}{k} \leqslant \sum_{k=1}^{m}\frac{k|x|}{k} = mx < 1$$
The second sum can be bounded as well using summation by parts.
Noting that $|S_n(x)| \leqslant \frac{1}{|\sin(x/2)|}$ and $|\sin(x/2)| \geqslant \frac{2}{\pi}\frac{x}{2} = \frac{x}{\pi} $ for $x \in (0,\pi)$ we have,
$$\left|\sum_{k=m+1}^{n}\frac{\sin kx}{k}\right| = \left|\frac{S_n(x)}{n} - \frac{S_m(x)}{m+1} + \sum_{k=m+1}^{n-1} S_k(x) \left(\frac{1}{k} - \frac{1}{k+1} \right)\right| \\ \leqslant \frac{2}{(m+1)|\sin(x/2)|} \\ \leqslant \frac{2\pi}{(m+1)x}\\\leqslant 2\pi$$
since $m = \lfloor 1/x \rfloor$ implies $(m+1)x \geqslant 1$.
You have to prove that your upper bound is valid. Use the fact that $e^{x} \geq x$ for $x >0$ (which follows from the Taylor expansion). Now $e^{n^{2}x^{2}} \geq e^{n^{2}a^{2}} \geq n^{2}a^{2}$ so $\frac 1 {e^{n^{2}x^{2}}} \leq \frac 1 {n^{2}a^{2}}$, now compare with the series $ \sum \frac 1 {n^{2}a^{2}}$ which is convergent.
Best Answer
We can prove that the series is not uniformly convergent on $[0,\infty)$ by showing there exists $\epsilon_0 > 0$ such that for any $N \in \mathbb{N}$ there are positive integers $q>p> N$ and $x_N\in [0,\infty) $ such that
$$\tag{*}\sum_{n=p}^q \frac{(-1)^{n-1}}{\sqrt{n}}\sin \frac{x_N}{n} \geqslant \epsilon_0$$
Taking $p = 4N+1$ and $q = 8N$ we have
$$\sum_{n= 4N+1}^{8N} \frac{(-1)^{n-1}}{\sqrt{n}}\sin \frac{x_N}{n} = \sum_{j=0}^{N-1}\sum_{k=1}^4\frac{(-1)^{4N+4j +k-1}}{\sqrt{4N+4j +k}}\sin \frac{x_N}{4N+4j+k} $$
The first step is to find $x_N$ such that if $n \in\{4N+1,\ldots,8N\}$ and $n \equiv 1,2,3 \text{ (mod }4)$, then $(-1)^{n-1}$ and $\sin \frac{x_N}{n}$ have the same sign and $(-1)^{n-1} \sin \frac{x_N}{n} > \frac{1}{\sqrt{2}}$.
Define $Q_N \in \mathbb{N}$ as $$Q_N =\frac{\prod_{n= 4N+1}^{8N}n}{2^N\prod_{j=0}^N(4N+4j+4)} \\=2(4N+1)(2N+2)(4N+3)(4N+5)(2N+3)(4N+7) \cdots (8N-3)(4N-1)(8N-1)$$
and $X_N = 2N\pi + Q_N \pi$.
If $k = 1,3$, then $n= 4N+4j+k$ is odd, $(-1)^{4N+4j +k-1}=1$, $\frac{Q_N}{n}=2l$ is even, and
$$\frac{X_N}{n} \in \left(\frac{\pi}{4} + 2l\pi, \frac{\pi}{2} + 2l\pi\right) \implies \frac{(-1)^{4N+4j +k-1}}{\sqrt{4N+4j +k}}\sin \frac{x_N}{4N+4j+k}> \frac{\frac{1}{\sqrt{2}}}{\sqrt{4N+4j +k}} $$
If $k = 2$ then $n= 4N+4j+k$ is even, $(-1)^{4N+4j +k-1}=-1$, $\frac{Q_N}{n}=2l+1$ is odd, and
$$\frac{X_N}{n} \in \left(\frac{3\pi}{4} + 2l\pi, \frac{3\pi}{2} + 2l\pi\right) \implies \frac{(-1)^{4N+4j +k-1}}{\sqrt{4N+4j +k}}\sin \frac{x_N}{4N+4j+k}> \frac{(-1)(-\frac{1}{\sqrt{2}})}{\sqrt{4N+4j +k}} $$
If $k = 4$ then $n= 4N+4j+k$ is even, $(-1)^{4N+4j +k-1}=-1$, $\frac{Q_N}{n}$ is not an integer, and since $\sin \theta < 1$ for any $\theta$,
$$\frac{(-1)^{4N+4j +k-1}}{\sqrt{4N+4j +k}}\sin \frac{x_N}{4N+4j+k}> \frac{-1}{\sqrt{4N+4j +k}}$$
Thus,
$$\sum_{k=1}^4\frac{(-1)^{4N+4j +k-1}}{\sqrt{4N+4j +k}}\sin \frac{x_N}{4N+4j+k} \\ > \frac{1/\sqrt{2}}{\sqrt{4N+4j+1}}+\frac{1/\sqrt{2}}{\sqrt{4N+4j+2}}+\frac{1/\sqrt{2}}{\sqrt{4N+4j+3}}-\frac{1}{\sqrt{4N+4j+4}}> \frac{3/\sqrt{2}-1}{\sqrt{4N+4j+4}},$$
and, using (*) we get
$$\sum_{n= 4N+1}^{8N} \frac{(-1)^{n-1}}{\sqrt{n}}\sin \frac{x_N}{n}> \sum_{j=0}^{N-1}\frac{3/\sqrt{2}-1}{\sqrt{4N+4j+4}}> N\cdot \frac{3/\sqrt{2}-1}{\sqrt{8N}}> \frac{3/\sqrt{2}-1}{\sqrt{8}}$$
Hence, with $\epsilon_0 = \frac{3/\sqrt{2}-1}{\sqrt{8}}$ condition (*) is satisfied for every $N$ (the Cauchy criterion for uniform convergence is violated) and the series is not uniformly convergent on $[0,\infty)$.
A similar argument applies in showing that convergence is not uniform for $x \in (-\infty,0]$.