Suppose $n\in \mathbb{N}_+$, $4n-2\leqslant\lambda \leqslant 4n$. How to prove $$\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{2k-1}\leqslant\frac12\int_0^{\pi/2}\frac{\sin \lambda x}{\sin x}dx\leqslant\sum_{k=1}^{2n-1}\frac{(-1)^{k-1}}{2k-1}\ ?$$
We know that forall $x\in (0,\pi)$,
\begin{align*} \frac{4}{\pi}\sum_{k=1}^{n}\frac{1}{2k-1}\sin(2k-1)x=\frac{2}{\pi}\int_0^x\frac{\sin 2nt}{\sin t} dt, \end{align*} or \begin{align*} \sum_{k=1}^{n}\frac{1}{2k-1}\sin(2k-1)x=\frac12\int_0^x\frac{\sin 2nt}{\sin t} dt. \end{align*} Place $x=\pi/2$, and replace $n$ by $2n,2n-1$ respectively, we have \begin{align*} \sum_{k=1}^{2n}\frac{(-1)^{k-1}}{2k-1}=&\frac12\int_0^{\pi/2}\frac{\sin 4nt}{\sin t} dt,\\ \sum_{k=1}^{2n-1}\frac{(-1)^{k-1}}{2k-1}=&\frac12\int_0^{\pi/2}\frac{\sin (4n-2)t}{\sin t} dt. \end{align*}
I wonder why it choose $\lambda\in [4n-2,4n]$. I have tried to prove that $$f(\lambda)=\int_0^{\pi/2}\frac{\sin \lambda x}{\sin x}d x$$ is decreasing, but without any result.
Since you've obtained $f(4n-2)=2S_{2n-1}$ and $f(4n)=2S_{2n}$, the conjecture you're trying to prove would rather be $2S_{2n}\leqslant f(\lambda)\leqslant 2S_{2n-1}$ for $4n-2\leqslant\lambda\leqslant 4n$. But it doesn't hold. The minima $\lambda_n\in[4n-2,4n]$ of $f(\lambda)$ are slightly less than $4n$ for $1\leqslant n\leqslant 5$ (at least): $$\begin{array}{c|c|c|c|} n&\lambda_n&f(\lambda_n)&f(4n)\\\hline 1&3.9063687163-&1.3307753943-&1.3333333333+\\ 2&7.9505549212+&1.4472480295+&1.4476190476+\\ 3&11.966609219-&1.4879093073-&1.4880230880+\\ 4&15.974835197-&1.5084872853-&1.5085359085+\\ 5&19.979821495+&1.5208947612+&1.5209198095-\\ \hline\end{array}$$ [An interesting question would be the asymptotics of $4n-\lambda_n$...]