The main problem (at least for me) is to prove that sum of the series $ f_\alpha(x)=\sum\limits_{k=1}^\infty\frac{\sin kx}{k^\alpha},\quad x\in[0,2\pi] $ is discontinuous for $\alpha\in(0,1]$.
Lemma 1. For $\alpha\in(0,1]$ we have $ \sum\limits_{k=1}^\infty\int\limits_{k-1/2}^{k+1/2}\frac{\sin(xt)}{k^\alpha}dt= \int\limits_{1/2}^\infty\frac{\sin(xt)}{t^\alpha}dt+\varphi(x) $ where $|\varphi(x)|\leq 2^{1-\alpha}$ for all $x\in[0,2\pi]$.
Proof. It is enough to show that difference between this sum and this integral is bounded by some constant. Now, we make estimation $ \left|\sum\limits_{k=1}^\infty\int\limits_{k-1/2}^{k+1/2}\frac{\sin(xt)}{k^\alpha}dt- \int\limits_{1/2}^\infty\frac{\sin(xt)}{t^\alpha}dt\right|= \left|\sum\limits_{k=1}^\infty\int\limits_{k-1/2}^{k+1/2}\left(\frac{\sin(xt)}{k^\alpha}- \frac{\sin(xt)}{t^\alpha}\right)dt\right|\leq $ $ \sum\limits_{k=1}^\infty\int\limits_{k-1/2}^{k+1/2}|\sin(xt)|\left|\frac{1}{k^\alpha}-\frac{1}{t^\alpha}\right|dt\leq \sum\limits_{k=1}^\infty\int\limits_{k-1/2}^{k+1/2}\left|\frac{1}{k^\alpha}-\frac{1}{t^\alpha}\right|dt= $ $ \sum\limits_{k=1}^\infty\left(\int\limits_{k-1/2}^{k}\left(\frac{1}{t^\alpha}-\frac{1}{k^\alpha}\right)dt+\int\limits_{k}^{k+1/2}\left(\frac{1}{k^\alpha}-\frac{1}{t^\alpha}\right)dt\right)= \sum\limits_{k=1}^\infty\left(\int\limits_{k-1/2}^{k}\frac{1}{t^\alpha}dt-\int\limits_{k}^{k+1/2}\frac{1}{t^\alpha}dt\right)\leq $ $ \sum\limits_{k=1}^\infty\left(\int\limits_{k-1/2}^{k}\frac{1}{(k-1/2)^\alpha}dt-\int\limits_{k}^{k+1/2}\frac{1}{(k+1/2)^\alpha}dt\right)\leq \sum\limits_{k=1}^\infty\left(\frac{1}{2(k-1/2)^\alpha}-\frac{1}{2(k+1/2)^\alpha}\right)=2^{\alpha-1} $
Lemma 2. For $\alpha\in(0,1]$ we have $ \sum\limits_{k=1}^\infty\frac{\sin kx}{k^\alpha}= \frac{x^\alpha}{2\sin(x/2)}\int\limits_{x/2}^\infty\frac{\sin y}{y^\alpha}+\frac{x\varphi(x)}{2\sin(x/2)} $ where $|\varphi(x)|\leq 2^{1-\alpha}$ for all $x\in[0,2\pi]$.
Proof. Note that $ \int\limits_{k-1/2}^{k+1/2}\sin(xt)dt= -\frac{1}{x}\cos(xt)\biggl|_{k-1/2}^{k+1/2}= \frac{2\sin(kx)\sin(x/2)}{x} $ so, $ \sin(kx)=\frac{x}{2\sin(x/2)}\int\limits_{k-1/2}^{k+1/2}\sin(xt)dt $ Hence from lemma 1 we conclude $ \sum\limits_{k=1}^\infty\frac{\sin kx}{k^\alpha}= \sum\limits_{k=1}^\infty\frac{x}{2k^\alpha\sin(x/2)}\int\limits_{k-1/2}^{k+1/2}\sin(xt)dt= \frac{x}{2\sin(x/2)}\sum\limits_{k=1}^\infty\int\limits_{k-1/2}^{k+1/2}\frac{\sin(xt)}{k^\alpha}dt= $ $ \frac{x}{2\sin(x/2)}\left(\int\limits_{1/2}^\infty\frac{\sin(xt)}{t^\alpha}dt+\varphi(x) \right) $ Making substitution $y=tx$ we get $ \sum\limits_{k=1}^\infty\frac{\sin kx}{k^\alpha}= \frac{x}{2\sin(x/2)}\left(\frac{1}{x^{1-\alpha}}\int\limits_{x/2}^\infty\frac{\sin y}{y^\alpha}dy+\varphi(x) \right)= \frac{x^\alpha}{2\sin(x/2)}\int\limits_{x/2}^\infty\frac{\sin y}{y^\alpha}dy+\frac{x\varphi(x)}{2\sin(x/2)} $
Corollary 3. For $\alpha\in(0,1]$ the function $f_\alpha$ is discontinuous at $0$.
Proof. Obviously $f_\alpha(0)=0$. Let $\alpha\in(0,1)$, then from fromula proved in lemma 2 we see that $ \lim\limits_{x\to +0}f_\alpha(x)=\lim\limits_{x\to +0}\left(\frac{x^\alpha}{2\sin(x/2)}\int\limits_{x/2}^\infty\frac{\sin y}{y^\alpha}dy+\frac{x\varphi(x)}{2\sin(x/2)}\right) $ Since $\varphi$ is bounded then the second term is bounded while the first tends to infinity. Hence the last limit is $\lim\limits_{x\to +0}f_\alpha(x)=+\infty$.
If $\alpha=1$, then $|\varphi(x)|\leq 1$ and since $ \lim\limits_{x\to+0}\frac{x}{2\sin(x/2)}=1, $ then $ \left|\frac{x\varphi(x)}{2\sin(x/2)}\right|<\frac{\pi}{3} $ for some $\delta_1>0$ and $x\in(0,\delta_1)$. Since $ \lim\limits_{x\to+0}\frac{x}{2\sin(x/2)}\int\limits_{x/2}^\infty\frac{\sin y}{y}dy= \int\limits_{0}^\infty\frac{\sin y}{y}dy=\frac{\pi}{2} $ then $ \frac{x}{2\sin(x/2)}\int\limits_{x/2}^\infty\frac{\sin y}{y}dy>\frac{2\pi}{5} $ for some $\delta_2>0$ and all $x\in(0,\delta_2)$. Thus for all $x\in(0,\min(\delta_1,\delta_2))$ we see that $ f_\alpha(x)=\frac{x^\alpha}{2\sin(x/2)}\int\limits_{x/2}^\infty\frac{\sin y}{y^\alpha}dy+\frac{x\varphi(x)}{2\sin(x/2)}>\frac{2\pi}{5}-\frac{\pi}{3}>0 $
In both cases $\lim\limits_{x\to+0}f_\alpha(x)\neq 0$. hence $f_\alpha$ is discontinuous at $0$.
Corollary 4. For $\alpha\in(0,1]$ the series $ \sum\limits_{k=1}^\infty\frac{\sin kx}{k^\alpha} $ doesn't converges uniformly on $[0,2\pi]$.
Proof. Assume that this series converges uniformly. Since this series is the sum of continuous functions and its converges uniformly, then its sum $f_\alpha$ must be continuous function on $[0,2\pi]$. This contradicts corallary 3, hence our series is not uniformly convergent on $[0,2\pi]$.
Remark 5. Despite the above, this series converges uniformly on $[\delta,2\pi-\delta]$ for all $\delta\in(0,\pi]$. You can use Dirichlet test to prove this.