Prove that if $\{a_{k}\}$ is a sequence of real numbers such that
$$\sum_{k=1}^{\infty} \frac{|a_{k}|}{k} = \infty$$ and
$$\sum_{n=1}^{\infty} \left( \sum_{k=2^n-1}^{2^n-1} k(a_k - a_{k+1})^2 \right)^{1/2} < \infty,$$ then
$$\int_{0}^{\pi} \left| \sum_{k=1}^{\infty} a_k \sin(kx) \right| \,dx = \infty.$$
My idea to prove Although the condition $$\lim_{k \rightarrow \infty} a_k=0$$ does not appear in the statement of the problem, by the well-known Cantor-Lebesgue theorem, this follows from the fact that the series $$\sum_{k=1}^{\infty} a_k \sin k x$$ is convergent almost everywhere. We note that for the application of this theorem, it would be sufficient if the series (2) were convergent on a set of positive measure. To make reference easier, we list the remaining conditions: $$\sum_{k=1}^{\infty} \frac{\left|a_k\right|}{k}=\infty$$
$$\sum_{n=1}^{\infty}\left(\sum_{k=2^{n-1}}^{2^n-1} k\left|\Delta a_k\right|^2\right)^{1 / 2}<\infty$$ where $$\Delta a_k:=a_k-a_{k+1} \quad(k=1,2, \ldots) .$$
From (4) it follows that the sequence $\left\{a_k\right\}$ has bounded variation, that is, $$\sum_{k=1}^{\infty}\left|\Delta a_k\right|<\infty$$
Really, by the Cauchy inequality,
\begin{aligned} \sum_{k=1}^{\infty}\left|\Delta a_k\right| & =\sum_{n=1}^{\infty} \sum_{k=2^{n-1}}^{2^n-1}\left|\Delta a_k\right| \\ & \leq \sum_{n=1}^{\infty}\left(2^{n-1} \sum_{k=2^{n-1}}^{2^n-1}\left|\Delta a_k\right|^2\right)^{1 / 2} \\ & \leq \sum_{n=1}^{\infty}\left(\sum_{k=2^{n-1}}^{2^n-1} k\left|\Delta a_k\right|^2\right)^{1 / 2} \end{aligned}
Consider the $n$ th partial sum of (2). By Abel's rearrangement, we obtain $$ \sum_{k=1}^n a_k \sin k x = \sum_{k=1}^n \tilde{D}_k(x) \Delta a_k + a_{n+1} \tilde{D}_n(x) $$ where $\tilde{D}_n(x)$ is the conjugate Dirichlet kernel: \begin{align*} \tilde{D}_n(x) &:= \sum_{k=1}^n \sin k x \\ &= \frac{\cos \frac{x}{2} - \cos \left(n + \frac{1}{2}\right) x}{2 \sin \frac{x}{2}} \quad (n=1,2, \ldots) . \end{align*}
Introduce the notation $$\bar{D}_n(x):=-\frac{\cos \left(n+\frac{1}{2}\right) x}{2 \sin \frac{x}{2}} \quad(n=0,1, \ldots) .$$
Then $$\tilde{D}_n(x)=\bar{D}_n(x)-\bar{D}_0(x) \quad\left(n=0,1, \ldots ; \tilde{D}_0(x)=0\right),$$
I think the following is a viable approach to this problem.
Set $$f(x) = 2 \sum_{k = 1}^\infty a_k \sin(kx).$$ My idea is to consider the following "model function" $g(x)$. For each positive integer $k$, define $$g(x) = \frac{a_{2^k} - a_1 + a_1 \cos(x/2)}{\sin(x / 2)}, \forall |x| \in [2^{-k-1}, 2^{-k}]$$ and $0$ when $|x| > 1/2$.
The crucial lemma is that $g$ approximates $f$ quite well in the $L^1$-sense. This idea is sort of reflected in what the OP wrote, but it is hard to implement in practice.
Lemma: We have $$\int_{-1/2}^{1/2} |f(x) - g(x)| dx < \infty.$$ Proof: Note that $$\sin(x/2) f(x) = \sum_{i = 1}^\infty a_i (\cos((i - 1/2) x) - \cos((i + 1/2) x)) = a_1 \cos(x / 2) + \sum_{i = 1}^\infty (a_{i + 1} - a_i) \cos((i + 1/2) x).$$ On the other hand, when $|x| \in [2^{-k-1}, 2^{-k}]$, we have $$\sin(x / 2) g(x) = a_0 \cos(x / 2) + \sum_{i = 1}^{2^k - 1} (a_{i + 1} - a_i).$$ So we have $$\sin(x/2) (f(x) - g(x)) = \sum_{i = 1}^{2^k - 1} (a_{i + 1} - a_i)(\cos((i + 1/2) x) - 1) + \sum_{i = 2^k}^{\infty} (a_{i + 1} - a_i) \cos((i + 1/2) x).$$ We now form functions $$d_k(x) = \frac{1}{\sin(x/2)} \cdot \begin{cases} \sum_{i = 2^k}^{2^{k + 1} - 1} (a_{i + 1} - a_i) \cos((i + 1/2) x), |x| \geq 2^{-k-1} \\ \sum_{i = 2^k}^{2^{k + 1} - 1} (a_{i + 1} - a_i) (\cos((i + 1/2) x) - 1), |x| < 2^{-k-1} \end{cases}.$$ Then for any $|x| \leq 1/2$, we have $$f(x) - g(x) = \sum_{k = 1}^\infty d_k(x).$$ Thus, we have the triangle inequality $$\int_{-1/2}^{1/2} |f(x) - g(x)| dx \leq \sum_{k = 1}^\infty \int_{-1/2}^{1/2} |d_k(x)| dx.$$ We now prove that
Claim: We have $$\int_{-1/2}^{1/2} |d_k(x)| dx \leq 1000 \left(\sum_{i = 2^{k}}^{2^{k + 1} - 1} i |a_{i + 1} - a_i|^2\right)^{1/2}.$$ Proof: We split the integral into two parts. For $|x| \geq 2^{-k-1}$, we use the Cauchy-Schwarz inequality $$\int_{|x| \in [2^{-k-1}, 1/2]} |d_k(x)| dx \leq \sqrt{\int_{|x| \in [2^{-k-1}, 1/2]} \sin(x/2)^{-2} dx} \cdot \sqrt{\int_{|x| \leq 1/2} | \sum_{i = 2^k}^{2^{k + 1} - 1} (a_{i + 1} - a_i) \cos((i + 1/2) x) |^2dx}$$ For the first integral, we have $$\sin(x / 2) \geq x/4$$ in the range of interest, so $$\int_{|x| \in [2^{-k-1}, 1/2]} \sin(x/2)^{-2} dx \leq \int_{|x| \geq 2^{-k-1}} \frac{16}{x^2} dx \leq 32 \cdot 2^{k + 1}.$$ For the second integral, we use orthogonality to obtain $$\int_{|x| \leq 2\pi} | \sum_{i = 2^k}^{2^{k + 1} - 1} (a_{i + 1} - a_i) \cos((i + 1/2) x) |^2dx = 2 \pi \sum_{i = 2^k}^{2^{k + 1} - 1} (a_{i + 1} - a_i)^2.$$ Combining these observations prove that $$\int_{|x| \in [2^{-k-1}, 1/2]} |d_k(x)| dx \leq 100 \cdot 2^{k / 2} \cdot \left(\sum_{i = 2^k}^{2^{k + 1} - 1} (a_{i + 1} - a_i)^2\right)^{1/2}.$$ An analogous Cauchy-Schwarz argument $$\frac{1}{\sin(x/2)}\sum_{i = 2^k}^{2^{k + 1} - 1} (a_{i + 1} - a_i) (\cos((i + 1/2) x) - 1) \leq \sqrt{\sum_{i = 2^k}^{2^{k + 1} - 1} (\cos((i + 1/2) x) - 1)^2 / \sin(x/2)^2} \sqrt{\sum_{i = 2^k}^{2^{k + 1} - 1} |a_{i + 1} - a_i|^2}$$ together with the estimate $(1 - \cos(i + 1/2) x) / \sin(x/2) \leq 2^{2k + 4} x$ deals with $|x| \leq 2^{-k-1}$ and proves the claim. Summing the claim over $k$, together with the somewhat mysterious second condition on $\{a_i\}$, proves the lemma. $\square$
By the lemma, it suffices to show that $$\int_{-1/2}^{1/2} |g(x)| dx = \infty.$$ One can show that $$\int_{-1/2}^{1/2} |a_1 (1 - \cos(x/2)) / \sin(x/2)| dx < \infty$$ So, if we define $$g'(x) = \frac{a_{2^k}}{\sin(x / 2)}, \forall |x| \in [2^{-k-1}, 2^{-k}].$$ Then it suffices to show that $$\int_{-1/2}^{1/2} |g'(x)| dx = \infty.$$ Note that $$\int_{|x| \in [2^{-k-1}, 2^{-k}]} |g'(x)| dx \geq \int_{|x| \in [2^{-k-1}, 2^{-k}]} \frac{2|a_{2^k}|}{x} dx \geq |a_{2^k}|.$$ Thus it suffices to show that $$\sum_{k = 1}^\infty |a_{2^k}| = \infty.$$ This follows from the fact that $$\sum_{k = 1}^\infty |a_{2^k}| \geq \sum_{k = 1}^\infty 2^{-k} \sum_{i = 2^k}^{2^{k + 1} - 1} |a_i| - \sum_{i = 1}^\infty |a_i - a_{i + 1}|.$$ By condition 1, the first sum is $\infty$. By condition 2 + Cauchy-Schwarz, the second sum is less than $\infty$. So we finish the proof. (Sorry this is brief, the formula rendering is becoming slow for me.)