Given $\epsilon>0$, let $\delta = \left( \frac \epsilon M\right)^{\frac1\alpha}$. Then $|x-y|<\delta$ implies $|f(x)-f(y)|\le M |x-y|^\alpha< M\delta^\alpha = M\frac \epsilon M=\epsilon$ as was to be shown.
Assume $\alpha>1$ and $f(x)\ne f(y)$. For $n\in \mathbb N$ let $x_k=x+\frac kn (y-x)$, $0\le k\le n$ (so that $x_0=x$, $x_n=y$). Then $|f(x_{k+1})-f(x_k)|\le M|x_{k-1}-x_k|^\alpha=M\left(\frac{|y-x|}n\right)^\alpha$ for $0\le k and hence $\tag1|f(y)-f(x)|\le \sum_{k=0}^{n-1}|f(x_{k+1})-f(x_k)|\le n M \left(\frac{|y-x|}n\right)^\alpha= M|y-x|^\alpha n^{1-\alpha}.$ As $1-\alpha<0$, the factor $n^{1-\alpha}$ gets arbitrarily small as $n\to\infty$, especially there exists $n$ such that $\tag2n^{1-\alpha}<\frac{|f(x)-f(y)|}{M|y-x|^\alpha}.$ (More explicitly, take $n>\left(\frac{M|y-x|^\alpha}{|f(x)-f(y)|}\right)^{\frac1{\alpha-1}}$). But (1) and (2) together imply $|f(y)-f(x)|<|f(y)-f(x)|$, a contradiction. Hence $f(x)\ne f(y)$ is not possible if $\alpha>1$.