I want to show that $$\zeta(z) = \frac{1}{z-1} + \gamma + O(z-1)$$
for $z \rightarrow 1 $. Why is it enough to show this statement for real $z = s \in (1,2)$? We know that $\zeta$ is meromorphic and has only one simple pole at $z=1$ but how does this help?
Now, for $s \in (1,2)$ we can write
$$\zeta(z) - \frac{1}{z-1} = \sum_{n=1}^{\infty} k^{-s} - \int_{1}^{\infty}t^{-s}dt =\sum_{k=1}^{\infty}\int_k^{k+1}(k^{-s} - t^{-s}) dt$$
and since $|k^{-s}-t^{-s}| = |\int_{k}^{t}su^{-s-1}du| \leq |s|k^{-s-1}$ we know that there exists a constant M such that
$$\sum_{k=1}^{\infty}\int_k^{k+1}(k^{-s} - t^{-s}) dt< M$$
for all $s \in (1,2)$. This implies the whole thing is uniform there and that we are allowed pull in limit, in the end we have:
$$\lim_{s \rightarrow 1}(\zeta(z) - \frac{1}{z-1}) = \sum_{k=1}^{\infty} \int_k^{k+1}(k^{-1}-t^{-1}) dt = \sum_{k=1}^{\infty}(k^{-1}-ln(k+1)+ln(k)) = \gamma$$.
I'm still a little confused with big O-notation, how does this imply our statement for general $z$?
$\newcommand{\d}{\,\mathrm{d}}$Before I discuss how the claim follows, I disagree with the assertion: "this implies the whole thing is uniform". There are two limit interchanges that need to be made to justify this proof in more detail:
$$\begin{align}\lim_{s\to1^+}\sum_{k=1}^\infty\int_k^{k+1}(k^{-s}-t^{-s})\d t&\overset{?!}{=}\sum_{k=1}^\infty\lim_{s\to1^+}\int_k^{k+1}(k^{-s}-t^{-s})\d t\\&\overset{?!}{=}\sum_{k=1}^\infty\int_k^{k+1}\lim_{s\to1^+}(k^{-s}-t^{-s})\d t\end{align}$$
And the bound: $$\exists M>0,\,\forall1<s<2:\quad\quad\sum_{k=1}^\infty\int_k^{k+1}(k^{-s}-t^{-s})\d t<M$$Is not strong enough to justify either of those limit interchanges. What we need is uniform convergence (if you want to prove it by "making everything uniform") - I need to know that, for all $\epsilon>0$ there is some $N\in\Bbb N$ with $\left|\sum_{k=n+1}^\infty\int_k^{k+1}(k^{-s}-t^{-s})\d t\right|<\epsilon$ for all $1<s<2$ and all $n\ge N$. That has not yet been shown. However, it is really quite easy to fix this using the same ideas: $$\begin{align}\forall n\in\Bbb N,\,\forall1<s<2:\quad\left|\sum_{k=n+1}^\infty\int_k^{k+1}(k^{-s}-t^{-s})\d t\right|&\le\sum_{k=n+1}^\infty|s|k^{-s-1}\\&\le\int_n^\infty st^{-s-1}\d t\\&=n^{-s}\\&\le n^{-1}\end{align}$$
So we can be confident the series uniformly converges. That justifies the first limit interchange! We then need to justify: $$\forall k\in\Bbb N:\quad\lim_{s\to1^+}\int_k^{k+1}(k^{-s}-t^{-s})\d t\overset{?!}{=}\int_k^{k+1}\lim_{s\to1^+}(k^{-s}-t^{-s})\d t$$This is also easy to justify by "making everything uniform" (or the dominated convergence theorem...). Note that: $$\begin{align}|k^{-s}-t^{-s}-(k^{-1}-t^{-1})|&=|(t^{-s}-t^{-1})+(k^{-1}-k^{-s})|\\&\le t^{-1}-t^{-s}+k^{-1}-k^{-s}\\&\le2(1-s^{-1})\end{align}$$For all admissible $k,t,s$ (justifying the final inequality with calculus). So the integrand uniformly converges on $[k,k+1]$ to $k^{-1}-t^{-1}$ and we can be happy. Note the convergence is actually uniform on $(1,\infty)$, which is stronger than what we need.
By the machinery of complex analysis. This has already been answered in the comments, I don't have much to add here.
I know $\zeta$, as function holomorphic on $\Bbb C\setminus\{-1\}$, must have a Laurent expansion at $1$. Examining such a Laurent series in light of the fact $(s-1)\zeta(s)$ converges as $s\to 1^+$ on the real axis shows the Laurent series must start with a $(s-1)^{-1}$ coefficient and show $\zeta(s)$ looks like $(s-1)^{-1}+\gamma+a_1(s-1)+a_2(s-2)^2+\cdots=(s-1)^{-1}+\gamma+O(s-1)$. There is simply no other possiblity: the analyticity is highly restrictive.
Note in the case of real analytic functions, we could have something like $(s-1)f(s)=\gamma+\sqrt{s-1}+\cdots$ which is not of the form $\gamma+O(s-1)$... so technically the proof is incomplete; it's the fact we deal with a meromorphic function that forces everything to work out nicely.