Observe that the left-hand side is the sum of two convergent series. Let $N\geq 2$ be an integer tending to infinity. Truncate the first series at the $N$-th term and the second series at the $2N$-th term. The resulting finite sum $S_N$ equals
\begin{align*}
S_N&=\sum_{2\leq n\leq N}\frac{\log 2}{n\log n\log 2n}+\sum_{2\leq n\leq 2N}\frac{(-1)^n}{n\log n}\\
&=\sum_{2\leq n\leq N}\left(\frac{1}{n\log n}-\frac{1}{n\log 2n}\right)+\sum_{\substack{2\leq n\leq 2N\\\text{$n$ even}}}\frac{1}{n\log n}
-\sum_{\substack{2\leq n\leq 2N\\\text{$n$ odd}}}\frac{1}{n\log n}\\
&=\sum_{2\leq n\leq N}\frac{1}{n\log n}-2\sum_{\substack{4\leq n\leq 2N\\\text{$n$ even}}}\frac{1}{n\log n}+\sum_{\substack{2\leq n\leq 2N\\\text{$n$ even}}}\frac{1}{n\log n}
-\sum_{\substack{2\leq n\leq 2N\\\text{$n$ odd}}}\frac{1}{n\log n}\\
&=\sum_{2\leq n\leq N}\frac{1}{n\log n}+\frac{1}{2\log 2}-\sum_{\substack{4\leq n\leq 2N\\\text{$n$ even}}}\frac{1}{n\log n}-\sum_{\substack{2\leq n\leq 2N\\\text{$n$ odd}}}\frac{1}{n\log n}\\
&=\frac{1}{\log 2}+\sum_{3\leq n\leq N}\frac{1}{n\log n}-\sum_{\substack{4\leq n\leq 2N\\\text{$n$ even}}}\frac{1}{n\log n}
-\sum_{\substack{2\leq n\leq 2N\\\text{$n$ odd}}}\frac{1}{n\log n}\\
&=\frac{1}{\log 2}+\sum_{3\leq n\leq N}\frac{1}{n\log n}-\sum_{3\leq n\leq 2N}\frac{1}{n\log n}\\
&=\frac{1}{\log 2}-\sum_{N<n\leq 2N}\frac{1}{n\log n}.
\end{align*}
Estimating the last sum crudely, it follows that
$$\frac{1}{\log 2}-\frac{1}{\log N}<S_N<\frac{1}{\log 2}.$$
In particular,
$$\lim_{N\to\infty} S_N=\frac{1}{\log 2}.$$
Of course this limit equals the left-hand side in the original question, hence we are done.