How would I show that $ \forall x \in \mathbb{R^+} \exists n \in \mathbb{N} \text{ s.t } n \leq x < n+1 $ via induction? How would I show that $ \forall x \in \mathbb{R^+} \exists n \in \mathbb{N}  \text{ s.t } n \leq x < n+1 $ via induction?
I have as follows,
I am not allowed to use the Archimedean property yet, but I am allowed to assume that for every positive real $x$ there exists an positive integer $n$ such that $n > x$. 
Let 
$P(n):= \exists n \in \mathbb{N^+}\text{ s.t } n \leq x < n+1 \text{ for some } x \in R$
$P(1)$ is true, we take $\sqrt 2.$
Assume the truth of $P(n)$
$P(n+1):=n+1 \leq x < (n+1) +1$
Since $N$ is closed under the operation of adding 1 and the assumption that for every positive real $x$ there exists a positive integer $n$ such that $n > x$, we have the right hand side of the inequality. How would I go about the left hand side?
EDIT
This is been forced, I can't seem to use induction to prove this, for otherwise it is fairly simple.
 A: Let us first examine the case $x>0$.
For every $N \in \Bbb N \setminus \{0\}$, consider the following statement:
$$P(N) = "\forall x \in [0,N) \ \exists n \in \Bbb N \ (n \le x < n+1)" .$$
Note that $P(1)$ is true, because if $x \in [0,1)$ then $0 \le x < 0+1$, so just pick $n=0$ and be done.
Assume $P(N)$ true and let us prove $P(N+1)$. If $x \in [0,N)$, then $x$ falls under the assumption for $P(N)$, and there is nothing to prove. If $x \in [N, N+1)$, this is equivalent to $N \le x < N+1$, so just pick $n = N$ and this is it.
Therefore, we have proved by induction that for every $x>0$ there exist $n \in \Bbb N \setminus \{0\}$ such that $n \le x < n+1$.
According to the latest edit of the original question the following part is no longer necessary, but I shall just leave it here.
Now, if $x \le 0$, then:


*

*if $x \in \Bbb Z \setminus (\Bbb N \setminus \{0\})$ then just choose $n=x$ to obtain $n = x < x+1 = n+1$, which is true;

*finally, if $x \in (-\infty, 0] \setminus \Bbb Z$ then there exist $n \in \Bbb N \setminus \{0\}$ such that $n \le -x < n+1$, so $-n \ge x > -n-1$. Since $x$ is not an integer, the equality $-n = x$ cannot happen, so let us drop it, while replacing the stronger inequality $x > -n-1$ with the weaker one $x \ge -n-1$, thus obtaining $-n-1 \le x < (-n-1) + 1$, so for $n$ just choose $-n-1$.


Note that nowhere have we used that for every $x$ we can find an $n$ with $n>x$. The only thing used was to make explicit what "belonging to an interval" means.
A: First, prove the existence of that $n$. 
We can do that defining $E:= \{k\in \mathbb{Z} : k\ge x-1\}$
We know that $E$ is not empty because $\forall x\in \mathbb{R}, \exists n\in \mathbb{N} : n>x$.
We have that $E$ is minored by $x-1$ so $E$ has a minimum, let's say $n$.
We have $n\in E$, so $\boxed{n>x-1}$. Since $n-1\notin E$ we have $n-1\le x-1$ if and only if $\boxed{n\le x}$.
Then, if you want to prove the unicity of that $n$
Let's give us $m, n\in \mathbb{Z}$ s.t. 
$n\le x< n+1$ and $m\le x< m+1$.
We can combine these inequations to have 
$n-m\le x-m< m+1-m = 1$
and 
$m-n\le x-n< n+1-n = 1$.
This yields $|m-n|<1$ but $m, n\in \mathbb{Z}$, so $m-n = 0$, i.e. $m=n$
