Liouville's Theorem: If $f(z)$ entire, bounded function, then is a constant.
Let $a,b$ be any 2 points in the Complex plane. Using the Cauchy Integral formula:
$$f(b)-f(a)=\frac{1}{2\pi i}\int_{|z|=R}\frac{f(z)dz}{(z-b)}-\frac{1}{2\pi i}\int_{|z|=R}\frac{f(z)dz}{(z-a)}$$ where R is the radius of a circle that properly contains $a,b$.
Therefore,
$$f(b)-f(a)=\frac{1}{2\pi i}\int_{|z|=R}\frac{f(z)(b-a)dz}{(z-b)(z-a)}$$
Since:
$$|\frac{f(z)(b-a)}{(z-b)(z-a)}|\leq \frac{M|b-a|}{(R-|b|)(R-|a|)}$$
Using ML formula:
$$|f(b)-f(a)|=\frac{1}{2\pi}|\int_{|z|=R}\frac{f(z)(b-a)dz}{(z-b)(z-a)}|\leq \frac{M|b-a|}{(R-|b|)(R-|a|)} R$$
Take $R\rightarrow\infty$.
Hence, $|f(a)-f(b)|=0 \Rightarrow f(a)=f(b)$
Is this proof Valid?
It's fine, other than that the final $2\pi$ shouldn't be there (it cancels with the one that you forgot…). I would have added a thing that you left implicit:$$\bigl|f(b)-f(a)\bigr|\leqslant\lim_{R\to\infty}\frac{M|b-a|}{(R-|b|)(R-|a|)}R=0,$$and therefore $f(a)=f(b)$.