It seems clear that $x$ is intended to be a fixed number in the interval $(0,1)$.
The answer to the least upper bound question depends on how one defines $\mathbb{N}$. If $0$ is included in $\mathbb{N}$, then the least upper bound is $1$. If $0$ is not included, then the least upper bound is $x$. That will be easy to prove, since $x^n\lt x^1$ for every $n\ge 2$. So the least upper bound is in fact a maximum.
The greatest lower bound is $0$, but not for the reason you described. Certainly $0\lt x^n$ for every $n\in \mathbb{N}$, which makes $0$ a lower bound. We want to show that there is no lower bound greater than $0$. So you need to show that for any $b\gt 0$, there is an $n$ such that $x^n \lt b$.
To do this, I suggest rewriting $x$ as $\frac{1}{1+y}$, where $y$ is positive. We want to show that for some $n$, we have $\frac{1}{(1+y)^n}\lt b$, or equivalently that there is an $n$ such that $(1+y)^n \gt 1/b$.
You can prove that result by observing that $(1 + y)^n \ge 1 + ny$ for $y \ge 0$ (It is true for $y \gt -1$.) This is the Bernoulli Inequality, which you may already have met. The Inequality can be proved by induction. Or else you can show using the Binomial Theorem that $(1 + y)^n \ge 1 + ny$.
Yes, you're right: if $A$ contained two or more points, $a, b \in A$ with, say, $a < b$, then $glb(A) \le a < b \le lub(A)$.
Here's a countable, bounded subset of the reals that does not contain its lub or glb:
$$
A = \left\{ - \frac n {n+1} \mid n \in \mathbb{N}\right\} \cup \left\{ \frac n {n+1} \mid n \in \mathbb{N}\right\}
$$
$A$ is $\{\dots, -\frac 3 4, -\frac 2 3, -\frac 1 2, 0, \frac 1 2, \frac 2 3, \frac 3 4, \dots \}$. Clearly, $glb(A) = -1$ and $lub(A) = 1$, but neither is a member of $A$.
Best Answer
Note that $A=\{\frac{4}{x}+1: x \geq 1\}$
So $\inf A=1$
Indeed assume that exists $s>0$ such that $\frac{4}{x}+1 \geq 1+s, \forall x \geq 1$
Then we would have that $x \leq \frac{4}{s},\forall x \geq 1$ which is a contradiction.
So for every $s>0$ exists $x \geq 1$ such that $\frac{4}{x}+1<1+s$ proving that
$\inf A=1$