For any $n \in \mathbb{N}$ there exists a minimal element in the set $ \{ x\in \mathbb{N}\mid n< x \}$ Problem from V. Zorich Analysis textbook:

For any $n \in \mathbb{N}$ there exists a minimal element in the set  $ \{ x\in \mathbb{N}\mid n< x \}$  namely $\min\{x \in \mathbb{N}|n<x\}=n+1$*

Proof
1\ We show, that the set E of such $n\in\mathbb{N}$ for which the statement is true   concide with $\mathbb{N}$
First we chech, that $1\in E$, i.e
$$\min \{x\in N| 1<x\}=2$$
Check this statement using the induction principle:
Let $M=\{x\in \mathbb{N}|(x=1)\vee (2\leq x\}$
By defenition М have $1\in M$. Further, if $x\in M$, then either $x=1$ and then $x+1=2\in M$,or $2\leq x$, then $2\leq(x+1)$ and again $(x+1)\in M$.
Therefore, $M=\mathbb{N}$ and, if $(x\neq1)\wedge(x\in \mathbb{N})$, then $2\leq x$ then indeed $\min \{x\in \mathbb{N}|1<x\}=2)$
We show then that if $n \in E$  then $(n+1) \in E$
We notice first that if $x\in \mathbb{N}|n+1<x$, then 
$$(x-1)=y\in\{y \in \mathbb{N}|n<y\}$$ because by proved all natuoal numbers are non less then 1, hence $(n+1)<x=>(x\neq1)$ and then $(x-1)=y\in\mathbb{E}$
Let now $n\in \mathbb{E}$ i.e $\min \{y\in \mathbb{N}|n<y\}=n+1$ then $x-1\geq y\geq n+1$ and $x\geq n+2$ Therefore 
$$(x\in\{x\in \mathbb{N}|n+1<x\})=>(x\geq n+2)$$ and consequently, $\min \{x\in\mathbb{N|n+1<x\}=n+2}$ and it means that $(n+1)\in \mathbb{E}$
Questions:
in the line
$M=\{x\in \mathbb{N}|(x=1)\vee (2\leq x\}$ Why $(2\leq x\}$?
$x-1\geq y\geq n+1$ How we came to this?
 A: 
Why $(2 \leq x)$?

The author's aim is to show that for any $x \in \mathbb{N}$ with $x \neq 1$ that $2 \leq x$. In doing so he proves that $2$ is the minimal element of the set $\{x \in \mathbb{N} \mid 1 \lt x\}$.
Firstly, he needs to prove that $M = \mathbb{N}$. That might look obvious from the definition of $M$ but remember that the $\leq$ relation, at this stage, has only been defined by a set of axioms. Some useful properties have been established but the only one involving actual numerals is $0 \lt 1$. So this proof doesn't assume that $1 \leq 2$ or $2 \leq 3$, for example.

$x-1 \geq y \geq n+1$ How we came to this?

I don't know which particular step you have trouble with so I'll try to explain that section of the proof.
He defines $x$ and $y$ by: $$x \in \{x \in \mathbb{N} \mid n+1 \lt x\}$$ $$y = x-1$$
He then proves that $x-1 \in \{y \in \mathbb{N} \mid n \lt y\}$ by showing that $x-1 \in \mathbb{N}$ and also using the implication, $n+1 \lt x \implies n \lt x-1$ (using properties $1^0$ in $2.1.2.e$: $(x \leq y) \wedge (z \lt w) \implies (x+z \lt y+w)$).
Now, by letting $n \in E$, this means $n+1 = \min\{y \in \mathbb{N} \mid n \lt y\}$. It was just shown that if $x \in \{x \in \mathbb{N} \mid n+1 \lt x\}$ then $x-1 \in \{y \in \mathbb{N} \mid n \lt y\}$.
Because $n+1$, by assumption, is the minimal element of this set, we must then have $n+1 \leq x-1$. (Note also that $x-1=y \implies x-1 \geq y$.)
Therefore, we have shown that $x-1 \geq y \geq n+1$.
Next, $x-1 \geq n+1 \implies x \geq n+2$, again using properties $1^0$ in $2.1.2.e$
This proves $$(x\in\{x\in \mathbb{N} \mid n+1<x\}) \implies (x\geq n+2)$$
which directly implies that $n+2 = \min\{x\in \mathbb{N} \mid n+1 \lt x\}$.
If anything is still not clear, let me know.
