Convergence is not uniform on $(-\pi,\pi)$ (although it is on compact subintervals).
To prove non-uniform convergence note that
$$2\sum_{n=1}^{\infty} (-1)^{n+1} \frac{\sin nx }{n} = -2\sum_{n=1}^{\infty} \cos n\pi \frac{\sin nx }{n} = -2 \sum_{n=1}^{\infty} \frac{\sin n(\pi+x) }{n} $$
However, taking $x_n = -\pi + \frac{\pi}{4n} \in (-\pi,\pi)$ we have for $n < k \leqslant 2n$ that $\frac{\pi}{4} < k (\pi+x_n) \leqslant \frac{\pi}{2}$ which implies $\frac{1}{\sqrt{2}} < \sin k (\pi+x_n) \leqslant 1$ and for all $n \in \mathbb{N}$,
$$\sup_{x \in (-\pi,\pi)}\left| \sum_{k = n+1}^{2n}\frac{\sin k(\pi+x) }{k} \right|\geqslant \sum_{k = n+1}^{2n}\frac{\sin k(\pi+x_n) }{k} > \frac{1}{\sqrt{2}}\ \sum_{k=n+1}^{2n} \frac{1}{k} > \frac{1}{\sqrt{2}} \cdot n \cdot \frac{1}{2n} = \frac{1}{2\sqrt{2}}$$
The LHS fails to converge to $0$ as $n \to \infty$ and the Cauchy criterion for uniform convergence is violated.
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You have taken a good first step in showing that $g_n(x) \to g(x)$ uniformly on the interval $[\delta,\pi]$ for any $\delta$ where $0 < \delta < \pi$.
However, as proved here, convergence is not uniform on $[0,\pi]$ and we can't use the standard theorem to conclude immediately that
$$\tag{*}\lim_{n \to \infty}\int_0^\pi g_n(x) dx = \int_0^\pi g(x) dx$$
Nevertheless, we can prove (*) holds using the additional fact that the sequence of partial sums is uniformly bounded -- that is, there exists $M > 0$ such that
$$|g_n(x)| = \left|\sum_{k=1}^n \frac{\sin kx}{k} \right|\leqslant M$$
for all $n \in \mathbb{N}$ and all $x \in [0,\pi]$. For a proof of this fact, which is not trivial to show, see here. Furthermore, because $g_n(x) \to g(x)$ pointwise, it follows that $|g(x)| \leqslant M$ for all $x \in [0,\pi]$.
We can write
$$\left|\int_0^\pi (g_n(x) - g(x)) \, dx \right| =\left|\int_0^\delta (g_n(x) - g(x)) \, dx + \int_\delta^\pi (g_n(x) - g(x)) \, dx \right| \\ \leqslant \int_0^\delta |g_n(x)| \, dx + \int_0^\delta |g(x)| \, dx + \left|\int_\delta^\pi (g_n(x) - g(x)) \, dx \right| $$
Using the bounds $|g_n(x)|, |g(x)| \leqslant M$ we find that the first and second integrals on the RHS are each bounded by $M \delta$. Choosing $\delta \leqslant \epsilon/(4M)$ we get
$$\left|\int_0^\pi (g_n(x) - g(x)) \, dx \right| \leqslant 2M\delta + \left|\int_\delta^\pi (g_n(x) - g(x)) \, dx \right| \leqslant \frac{\epsilon}{2} + \left|\int_\delta^\pi (g_n(x) - g(x)) \, dx \right| $$
Since $g_n \to g$ uniformly on $[\delta,\pi]$, it follows that $\int_\delta^\pi g_n(x) \to \int_\delta^\pi g(x) \, dx$ and given $\epsilon > 0$ there exists $N$ such that for all $n > N$ we have $\left|\int_\delta^\pi (g_n(x) - g(x)) \, dx \right| < \epsilon/2$.
Therefore, for all $n > N$ we have
$$\left|\int_0^\pi g_n(x)\, dx -\int_0^\pi g(x)\, dx\right|= \left|\int_0^\pi (g_n(x) - g(x)) \, dx \right| \leqslant \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon,$$
and (*) is true.
Best Answer
You have shown correctly with the Dirichlet test that $(g_n)$ converges uniformly on the interval $[\delta, \pi]$ for any $\delta > 0$. This relies on the fact that $\left|\sum_{k=1}^n \sin kx \right| \leqslant 1 / \sin (\delta/2)$ is uniformly bounded for all $n \in \mathbb{N}$ and $x \in [\delta,\pi]$.
However, if $\delta = 0$ the partial sums are not bounded and the Dirichlet test is not applicable. This is a clue that convergence is not uniform on $[0,\pi]$.
If the convergence were uniform, then by the Cauchy criterion for any $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that for all $m > n \geqslant N$ and all $x \in [0,\pi]$ we would have
$$\tag{1}\left|\sum_{k=n+1}^{m}\frac{\sin kx }{k} \right| < \epsilon$$
We can show that this criterion is violated and the convergence is not uniform. Take $\epsilon = 1/(4\sqrt{2})$ and, for any integer $N$, no matter how large, choose $n = N$, $m = 2n$ and $x_n = \pi/(4n) \in [0,\pi]$. For any $k > n$, we have $kx_n > \pi/4$ and $\sin kx_n > \sin (\pi/4) = 1/\sqrt{2}$.
Hence,
$$\tag{2}\left|\sum_{k=n+1}^{2n}\frac{\sin kx_n }{k} \right| > \frac{1}{\sqrt{2}}\sum_{k=n+1}^{2n}\frac{1 }{k} > \frac{1}{\sqrt{2}}\cdot n \cdot \frac{1}{2n} = \frac{1}{2\sqrt{2}} > \epsilon,$$
and we have a contradiction to condition (1).