Approximating by incrementing positive integers

There is no such $\alpha$.

If $\alpha\in\mathbb Q$, there is $n$ such that $\mu(n,\alpha)=0$, thus $\mu(n+1,\alpha)<\mu(n,\alpha)$ is impossible.

If $\alpha\notin\mathbb Q$, a classical result in Diophantine approximation says that there are infinitely many $n$ such that $$\mu(n,\alpha)<\frac1{\sqrt5n^2}.$$ If then $$\mu(n+1,\alpha)<\mu(n,\alpha),$$ let $a/n$ and $b/(n+1)$ be the respective closest approximations of $\alpha$. We have $$\left|\frac an-\frac b{n+1}\right|<\frac2{\sqrt5n^2}<\frac1{n(n+1)},$$ while $$\left|\frac an-\frac b{n+1}\right|=\frac{|a(n+1)-bn|}{n(n+1)}\ge\frac1{n(n+1)}$$ unless $a(n+1)=bn$, i.e., the approximating fractions are $0$ or $1$. Since this happens for infinitely many $n$, this is impossible for $\alpha\in(0,1)$.

No. If $\alpha$ is rational, set $n$ to the denominator of $\alpha$. Otherwise set $n$ to the denominator of the third convergent to $\alpha$. In both cases, we get $\mu(n+1,\alpha)>\mu(n,\alpha)$.