Prove that $\bigcup_{x\in (1,\infty)}[\frac{1}{x},x)=(0, \infty)$

Prove that it holds.

Is my proof correct:

First I would define the union:

$$S = \Bigl\{y\in\mathbb{R}\mid\exists x \in(1,\infty);y\in\Bigl[\frac{1}{x},x\Bigr)\Bigr \}$$

i)

$$S \subset (0, \infty)$$

proof: There exists $x$ such that $\bigl[\frac{1}{x},x\bigr)\in(0,\infty)$, let's say $x = 2$

ii) $$(0, \infty) \subset S$$

proof:

Let's say that $z \in (0, \infty)$, we can pick $z = \frac{x}{2} \implies \forall z \in [\frac{1}{x},x) \implies \forall z \in S$

Is my proof correct ?

You're misusing quantifiers where you write:

$z=\frac{x}{2},\,\forall z\in[1/x,x)\implies\forall z\in S$

This sentence makes little sense. Pause, and ask what does "Implies for all $z$ in $S$" really mean? Implies... what?

You cannot use merely a single element to prove subset; that just proves intersection. For example, if $A=\{1,2,3\},\,B=\{3,4,5\}$, it is hopefully clear that $A\subset B$ is false, but $3\in A,\,3\in B$, so your logic:

$x=2$ has $[1/x,x)\in(0,\infty)$

Doesn't work. Moreover, you are misusing the "in" relationship; $[1/x,x)$ is not in $(0,\infty)$, as every member of $(0,\infty)$ is a distinct real number, whereas $[1/x,x)$ is an interval, not a real number. The correct symbol to capture the intuitive idea that it is "in" is subset; $[1/x,x)\subset(0,\infty),\,x\gt0$ is definitely true.

Onto the proof:

$A\subset B\iff\forall x\in A,\,x\in B$, by axiomatic definition. Let's see: $$\begin{align}y\in S\implies\exists x\gt1:y\in[1/x,x)&\implies\frac{1}{x}\le y\lt x\implies0\lt y\lt\infty\\&\implies y\in(0,\infty)\end{align}$$So we have $S\subset(0,\infty)$.

Now let's check the other direction: $$y\in(0,\infty)\implies0\lt y\lt\infty$$Recall that to be in $S$, we need an $x\gt 1$. Assume $y\ge1$ for the moment, and consider $x=y+1\gt1$: $$\frac{1}{x}\lt y\lt x\implies y\in[1/x,x)\implies y\in S$$Now if $0\lt y\lt 1$, we can instead consider $x=\frac{1}{y}\gt1$, and still have that: $$\frac{1}{x}\le y\lt x\implies y\in[1/x,x)\implies y\in S$$So $\forall y\in(0,\infty),\,y\in S,\,(0,\infty)\subset S$. We are done, as: $$S\subset(0,\infty)\subset S\implies S=(0,\infty)$$