Your proof is confusingly written.
Problem 1:
if it is a supremum, then $\exists x \in [a,b)$ s.t. $x+\epsilon>b\geq x$ for some $\epsilon>0$ (epsilon characterization of the supremum)
This is not the epsilon characterization of the supremum. The characterization says that:
$s$ is a supremum of $A$ if, for every $\epsilon > 0$, there exists some $x\in A$ such that $s-\epsilon < a$.
This is completely different from what you wrote. Most importantly, the definition of supremum says something is true for every epsilon, not "for some $\epsilon > 0$ which is what you wrote.
Problem 2:
You wrote "$\forall x \in b$", which makes no sense. $b$ is not a set, so $x\in b$ is nonsensical.
My reasoning is too long to write in the comment. So I write my own answer here.
$(1)$ To show that $0$ is the infimum.
$(i)$ $n^3+1 \gt 0$ and $n^4+16 \gt 0 \implies \frac{n^3+1}{n^4+16} \gt 0$.
$\therefore 0$ is a lower bound.
$(ii)$ For any $\varepsilon \gt 0$, we reason as follows:
$$\frac{n^3+1}{n^4+16} \lt \frac{n^3+n^3}{n^4}=\frac{2}{n}$$
So if we can find $n$ such that $\frac{2}{n} \lt \varepsilon$, then we are done. But this is easy, for this is equivalent to $n \gt \frac{2}{\varepsilon}.$
Formally we can write the proof as follows:
For any $\varepsilon \gt 0$, take $n_0=\lfloor \frac{2}{\varepsilon} \rfloor +1 $, then $n_0 \gt \frac{2}{\varepsilon}$ and hence $\varepsilon \gt \frac{2}{n_0}$.
Therefore
$$\frac{n_0^3+1}{n_0^4+16} \lt \frac{n_0^3+n_0^3}{n_0^4}=\frac{2}{n_0} \lt \varepsilon$$
$(i)$ and $(ii)$ prove that $0$ is the infimum.
$$$$
$(2)$ To prove that $\frac{28}{97}$ is the supremum,
we prove that $\frac{28}{97}$ is one of the terms and $$\frac{n^3+1}{n^4+16} \leq \frac{28}{97} \;\;\; \forall n \in \mathbb N.$$
Let $a_n=\frac{n^3+1}{n^4+16}$ and put $n=1, 2, 3, 4,$ we find that
$$a_1=\frac{2}{17}, a_2=\frac{9}{32}, a_3=\frac{28}{97}, a_4=\frac{65}{272}.$$
We see that $a_3=\frac{28}{97}$ is the largest of the the first $4$ terms.
Next we prove that $$\frac{n^3+1}{n^4+16} \leq\frac{28}{97} \;\; \forall n \geq 5. $$
This can be done as follows:
$\forall n \geq 5,$
\begin{align}
\frac{n^3+1}{n^4+16} & \lt \frac{n^3+1}{n^4} \\
& = \frac{1}{n}+\frac{1}{n^4} \\
& \leq \frac{1}{5}+\frac{1}{5^4} \\
& = \frac{126}{625} \\
& \lt \frac{28}{97}
\end{align}
This completes the proof that $\frac{28}{97}$ is the supermum.
Best Answer
We see that if $\frac{n-1}{n+1}>1$ and $n\in\mathbb Z^{+}$, then we get a contradiction: $n<-1.$
Thus we have,
$$0≤\frac{n-1}{n+1}≤1$$
Now, I will show that for any $0<\varepsilon<1$ there exist $n\in\mathbb Z^{+}$, such that
$$\frac{n-1}{n+1}>1-\varepsilon$$
Algebra tells us, we have
$$n>\frac 2\varepsilon-1,\thinspace\text {where}\thinspace \thinspace n\in\mathbb Z^{+}$$
To be more precise, we can take
$$n=\left\lfloor\frac 2\varepsilon\right\rfloor+1$$
Finally we conclude that, if $n=\left\lfloor\frac 2\varepsilon\right\rfloor+1$ then:
$$1-\varepsilon<\frac{n-1}{n+1}≤1,\thinspace \forall \varepsilon \in (0,1)$$
This means,
$$\sup\left\{\frac {n-1}{n+1},\thinspace n\in\mathbb Z^{+}\right\}=1.$$