Uniform convergence of $f: \mathbb{R} \rightarrow \mathbb{R} $ Where $f(\sigma) = \sum_{n=1}^{\infty}\mu(n)/n^\sigma$ Here $\mu(n)$ is Möbius function.
Note: this question is substantially changed after comments from Daniel and Reuns. The original question did not make any sense. The domain of function is changed to real numbers because, initially I was conflating between boundedness and convergence. I had wrongly assumed 
$\sum_{n=1}^{\infty} \mu(n)/n^s$ was bounded on the like $s = 1+ it$. It is not. as a result, the answer may seem disconnected.
There are two approaches one can take to find where $\sum_{n=1}^{\infty} \mu(n)/n^\sigma$ is uniformly convergent.
1) Because absolute convergence implies uniform convergence, the series of functions must be uniformly convergent for interval $\sigma \ge 1+\epsilon$. The $\epsilon$ is needed to make the domain compact - which is needed for uniform continuity.
2) Further, the norm $| f_n - f | < \epsilon $ where $\epsilon$ does not depend on $\sigma$ according to the following from Titchmarsh:

Does this make sense? 
 A: There is something that you are missing but it is hard to tell what. 

Under the RH, $\sum_n \mu(n) n^{-s}$ converges for $\Re(s) > 1/2$. $\tag{1}$

Let $\Re(s)>a>1/2$.
By partial summation $(1)$ implies that it is $= \sum_n (\sum_{m\le n} \mu(m)m^{-a})(n^{a-s}-(n+1)^{a-s})$ which converges absolutely and hence $(1)$ converges locally uniformly on $\Re(s) >1/2$, uniformly on $\Re(s)>a, |arg(s-a)|\le \pi/2-r$. We know that it doesn't converge uniformly on $\Re(s) > a$ because $1/\zeta(s)$ is unbounded on $\Re(s)=1$ (it follows from the Euler product and that the $\log p$ are $\Bbb{Q}$-linearly independent).
The proof of $(1)$ is in the last chapter of Titchmarsh, it is based on the same argument of shifting to the left $$\int_{2-iT}^{2+iT} \frac{x^{z-s}}{(z-s)\zeta(z)}dz=2i\pi \sum_{n\le x}\mu(n)n^{-s}+O(x^{2-s+\epsilon}/T)$$ $$=2i\pi Res(\frac{x^{z-s}}{(z-s)\zeta(z)})+\int_{a-iT}^{a+iT} \frac{x^{z-s}}{(z-s)\zeta(z)}dz\pm \int_{a\pm iT}^{2\pm iT} \frac{x^{z-s}}{(z-s)\zeta(z)}dz$$
As in the proof of the PNT, we need a bound for $\frac1{\zeta(s)}$ which will be $$\frac1{\zeta(s)}=O(t^\epsilon)$$ (it follows from $\log \zeta(2+it)=O(1)$, the density of zeros and $ (\log \zeta(s))'' = \sum_\rho \frac1{(s-\rho)^2} = o( \log s)$)
For the vertical part we obtain $\int_{a-iT}^{a+iT} \frac{x^{z-s}}{(z-s)\zeta(z)}dz=O(T^\epsilon x^{a-s})$. 
We'll take $T=x^3$ so that $O(x^{2-s+\epsilon}/T)=o(1),\int_{a\pm iT}^{2\pm iT} \frac{x^{z-s}}{(z-s)\zeta(z)}dz=o(1)$ and the result follows
$$2i\pi \sum_{n\le x}\mu(n)n^{-s}=2i\pi Res(\frac{x^{z-s}}{(z-s)\zeta(z)})+o(1)$$
