My attempt:
Given any $\epsilon > 0$, \begin{align} \sup_{|z| \leq R}\, \biggl\lvert \frac{nz}{n+1} +\frac{3}{n} - z \biggr\rvert &= \sup_{|z| \leq R}\, \biggl\lvert \frac{nz-nz-z}{n+1} + \frac{3}{n} \biggr\rvert \\ &= \sup_{|z| \leq R}\, \biggl\lvert \frac{-z}{n+1}+\frac{3}{n} \biggr\rvert \\ &\leq \frac{R}{n+1} + \frac{3}{n} \\ &< \frac{3}{n}. \end{align} Choose $N(\epsilon) = \bigl\lceil \frac{3}{\epsilon} \bigr\rceil$. Then $\frac{3}{n} < \epsilon$. Hence, proved.
I want to make sure that my proof is valid and also will it matter if we take the ceiling function of $\frac{\epsilon}{3}$ or not.
That is not correct, because you wrote that $\frac R{n+1}+\frac3n<\frac3n$, which is false.
You have$$\frac R{n+1}+\frac3n<\frac Rn+\frac3n=\frac{R+3}n.$$ So, take$$N(\varepsilon)=\left\lceil\frac{R+3}\varepsilon\right\rceil.$$