Infimum of $\{\frac{11}{n + 3} : n \in \mathbb{N}\}$
Your idea that you describe is perfectly OK.
If you set $n=\lceil\frac{12}{\epsilon}\rceil$, then you know that $n\geq \frac{12}{\epsilon}$.
From that inequality, you can then deduce the following:
$$n > \frac{11}{\epsilon}\\ n\epsilon > 11\\ (n+3)\epsilon > 11 \epsilon > \frac{11}{n+3}$$
Note that every line above follows from the previous line, so it proves that your choice of $n$ satisfies the condition $\frac{11}{n+3}<\epsilon$.
Since $\epsilon$ was arbitrary, you can therefore conclude that
$$\forall \epsilon \exists n: \frac{11}{n+3}<\epsilon$$
which of course proves that $0$ is the largest possible lower bound (i.e., the infimum) of your set.