[proofplan] Write $s=\sigma+it$ and choose the Euler--Maclaurin truncation height $N$ comparable to $T:=|t|$. The truncated Dirichlet sum is bounded by comparing $n^{-\sigma}$ with $n^{-\delta}$, which avoids any singular constant as $\sigma\to1$. The pole term is controlled by $|s-1|\ge |t|\ge1$, and the Euler--Maclaurin remainder estimates are stated with the logarithmic factor that appears after differentiation; this logarithm is absorbed into the weaker power $T^{2-\delta}$. [/proofplan] [step:Choose a truncation height comparable to the imaginary part] Let $s=\sigma+it \in \mathbb{C}$, where $\sigma=\operatorname{Re}(s)$ and $t=\operatorname{Im}(s)$. Define \begin{align*} T := |t| \geq 1, \qquad N := \lceil T\rceil \in \mathbb{N}. \end{align*} Then \begin{align*} T \leq N \leq T+1 \leq 2T. \end{align*} Throughout the proof, constants denoted by $C_{\delta,A}>0$ may change from line to line but depend only on $\delta$ and $A$. Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$. We use the one-term Euler--Maclaurin expansion obtained from summation by parts with the first periodic Bernoulli function. Let $B_1: [N,\infty)\to\mathbb{R}$ denote the bounded periodic function $B_1(x)=\{x\}-1/2$, where $\{x\}=x-\lfloor x\rfloor$. For $\sigma\geq\delta$ and $s\neq1$, it gives \begin{align*} \zeta(s) = \sum_{n=1}^{N} n^{-s} + \frac{N^{1-s}}{s-1} + R_N(s), \end{align*} where \begin{align*} R_N(s)=-\frac{1}{2}N^{-s}-s\int_N^\infty B_1(x)x^{-s-1}\,d\mathcal{L}^1(x). \end{align*} Since $|B_1(x)|\leq1/2$ and $\sigma\geq\delta$, the improper integral is absolutely convergent and \begin{align*} |R_N(s)| &\leq \frac{1}{2}N^{-\sigma}+\frac{|s|}{2}\int_N^\infty x^{-\sigma-1}\,d\mathcal{L}^1(x) \\ &\leq C_{\delta,A}(1+|s|)N^{-\sigma}. \end{align*} Differentiating this identity under the absolutely convergent improper integral gives \begin{align*} \zeta'(s) = -\sum_{n=1}^{N}(\log n)n^{-s} - \frac{N^{1-s}\log N}{s-1} - \frac{N^{1-s}}{(s-1)^2} + R_N'(s), \end{align*} where \begin{align*} R_N'(s)=\frac{1}{2}(\log N)N^{-s}-\int_N^\infty B_1(x)x^{-s-1}\,d\mathcal{L}^1(x)+s\int_N^\infty B_1(x)(\log x)x^{-s-1}\,d\mathcal{L}^1(x). \end{align*} The same bound on $B_1$ and the estimates \begin{align*} \int_N^\infty x^{-\sigma-1}\,d\mathcal{L}^1(x)&\leq \delta^{-1}N^{-\sigma}, \\ \int_N^\infty (\log x)x^{-\sigma-1}\,d\mathcal{L}^1(x)&=N^{-\sigma}\left(\frac{\log N}{\sigma}+\frac{1}{\sigma^2}\right) \leq C_\delta(1+\log N)N^{-\sigma} \end{align*} give \begin{align*} |R_N'(s)|\leq C_{\delta,A}(1+|s|)(1+\log N)N^{-\sigma}. \end{align*} [/step] [step:Bound the truncated Dirichlet sum] For the finite Dirichlet sum, the triangle inequality gives \begin{align*} \left|\sum_{n=1}^{N}n^{-s}\right| \leq \sum_{n=1}^{N} n^{-\sigma}. \end{align*} If $\delta\leq \sigma < 1$, then $n^{-\sigma}\leq n^{-\delta}$ for every $1\leq n\leq N$. Comparison with the integral of the decreasing function $x\mapsto x^{-\delta}$ gives \begin{align*} \sum_{n=1}^{N} n^{-\sigma} \leq \sum_{n=1}^{N}n^{-\delta} \leq 1+\int_1^N x^{-\delta}\,d\mathcal{L}^1(x) = 1+\frac{N^{1-\delta}-1}{1-\delta} \leq C_{\delta,A} N^{1-\delta}, \end{align*} where the constant is finite because $0<\delta<1$. If $\sigma\geq 1$, then $n^{-\sigma}\leq n^{-1}$ for every $1\leq n\leq N$, so \begin{align*} \sum_{n=1}^{N} n^{-\sigma} \leq 1+\int_1^N x^{-1}\,d\mathcal{L}^1(x) = 1+\log N. \end{align*} Since $0<\delta<1$, the elementary bound $1+\log N\leq C_{\delta,A}N^{1-\delta}$ holds for all $N\in\mathbb{N}$. Hence, uniformly for $\delta\leq\sigma\leq A$, \begin{align*} \left|\sum_{n=1}^{N}n^{-s}\right| \leq C_{\delta,A}N^{1-\delta}. \end{align*} Since $N\leq 2T$, this gives \begin{align*} \left|\sum_{n=1}^{N}n^{-s}\right| \leq C_{\delta,A}T^{1-\delta}. \end{align*} [guided] The finite sum is estimated without cancellation. Since $|n^{-s}|=n^{-\sigma}$, the triangle inequality gives \begin{align*} \left|\sum_{n=1}^{N}n^{-s}\right| \leq \sum_{n=1}^{N}n^{-\sigma}. \end{align*} The worst case in the strip occurs near the left edge $\sigma=\delta$. For $\delta\leq\sigma<1$, we avoid dividing by $1-\sigma$ by first using the pointwise comparison $n^{-\sigma}\leq n^{-\delta}$ for $n\geq1$. Then comparison with the integral of $x^{-\delta}$ gives \begin{align*} \sum_{n=1}^{N} n^{-\sigma} \leq \sum_{n=1}^{N}n^{-\delta} \leq 1+\int_1^N x^{-\delta}\,d\mathcal{L}^1(x) = 1+\frac{N^{1-\delta}-1}{1-\delta} \leq C_{\delta,A}N^{1-\delta}. \end{align*} The constant is finite because $0<\delta<1$, and it is independent of how close $\sigma$ is to $1$. For $\sigma\geq1$, each summand satisfies $n^{-\sigma}\leq n^{-1}$, so the harmonic estimate gives \begin{align*} \sum_{n=1}^{N}n^{-\sigma} \leq 1+\int_1^N x^{-1}\,d\mathcal{L}^1(x) = 1+\log N. \end{align*} The corrected hypothesis $0<\delta<1$ is exactly what makes this logarithm harmless: there is a constant $C_{\delta,A}>0$ such that $1+\log N\leq C_{\delta,A}N^{1-\delta}$ for every $N\in\mathbb{N}$. Therefore \begin{align*} \left|\sum_{n=1}^{N}n^{-s}\right| \leq C_{\delta,A}N^{1-\delta} \leq C_{\delta,A}T^{1-\delta}, \end{align*} because $N\leq2T$. [/guided] [/step] [step:Control the pole term and Euler--Maclaurin remainder] Since $|s-1|\geq |\operatorname{Im}(s-1)|=|t|=T\geq1$ and $|N^{1-s}|=N^{1-\sigma}$, \begin{align*} \left|\frac{N^{1-s}}{s-1}\right| \leq T^{-1}N^{1-\sigma} \leq N^{1-\sigma} \leq C_{\delta,A}N^{1-\delta} \leq C_{\delta,A}T^{1-\delta}. \end{align*} Also, because $\sigma\leq A$ and $T\geq1$, \begin{align*} |s| \leq |\sigma|+|t| \leq A+T \leq (A+1)T. \end{align*} The Euler--Maclaurin remainder estimate therefore gives \begin{align*} |R_N(s)| \leq C_{\delta,A}(1+|s|)N^{-\sigma} \leq C_{\delta,A}T N^{-\sigma} \leq C_{\delta,A}T^{1-\sigma} \leq C_{\delta,A}T^{1-\delta}. \end{align*} Combining this with the preceding step yields \begin{align*} |\zeta(s)|\leq C_{\delta,A}T^{1-\delta}. \end{align*} [/step] [step:Bound the differentiated finite sum] For the differentiated finite sum, the triangle inequality gives \begin{align*} \left|\sum_{n=1}^{N}(\log n)n^{-s}\right| \leq \sum_{n=1}^{N}(\log n)n^{-\sigma}. \end{align*} Since $\log n\leq \log N$ for $1\leq n\leq N$, the estimate from the finite sum gives \begin{align*} \sum_{n=1}^{N}(\log n)n^{-\sigma} \leq (\log N)\sum_{n=1}^{N}n^{-\sigma} \leq C_{\delta,A}N^{1-\delta}\log N. \end{align*} Because $N\leq2T$ and $T\geq1$, after increasing the constant $C_{\delta,A}$ the elementary inequality $\log(2T)\leq C_{\delta,A}T$ gives \begin{align*} \left|\sum_{n=1}^{N}(\log n)n^{-s}\right| \leq C_{\delta,A}T^{2-\delta}. \end{align*} [/step] [step:Control the differentiated pole and remainder terms] The first differentiated pole term satisfies \begin{align*} \left|\frac{N^{1-s}\log N}{s-1}\right| \leq T^{-1}N^{1-\sigma}\log N \leq C_{\delta,A}T^{1-\delta}, \end{align*} where we used $N\leq2T$ and $\log(2T)\leq C_{\delta,A}T$ for $T\geq1$. In particular this is at most $C_{\delta,A}T^{2-\delta}$. The second differentiated pole term satisfies \begin{align*} \left|\frac{N^{1-s}}{(s-1)^2}\right| \leq T^{-2}N^{1-\sigma} \leq C_{\delta,A}T^{1-\delta} \leq C_{\delta,A}T^{2-\delta}, \end{align*} using $T\geq1$ in the last inequality. Finally, the differentiated remainder estimate, $|s|\leq(A+1)T$, $N\leq2T$, and $\log(2T)\leq C_{\delta,A}T$ give \begin{align*} |R_N'(s)| \leq C_{\delta,A}(1+|s|)(1+\log N)N^{-\sigma} \leq C_{\delta,A}T^2N^{-\sigma} \leq C_{\delta,A}T^{2-\sigma} \leq C_{\delta,A}T^{2-\delta}. \end{align*} Substituting these three bounds and the differentiated finite-sum bound into the differentiated Euler--Maclaurin formula gives \begin{align*} |\zeta'(s)|\leq C_{\delta,A}T^{2-\delta}. \end{align*} Since $T=|t|=|\operatorname{Im}(s)|$, the two displayed inequalities are exactly the claimed bounds. [/step]