For every $n$ there exists $m$ such that $m/n$ is an upper bound but $(m-1)/n$ is not This is a problem discussed in Analysis 1 by Terence Tao.

$E$ is a non-empty set of Real numbers. $ n\geq 1$, $L,K$ are two integers such that $L<K$. Let $\frac{L}{n}$ is not an upper bound of $E$ but $\frac{K}{n}$ is an upper bound. It is required to prove that:
There exists an integer $m$ where $L<m \leq K$ such that $\frac{m}{n}$ is an upper bound for $E$ but  $\frac{m-1}{n}$ is not.

Suppose $E=(0,1)$ and $n=2$. Let $L=1 ,K=3$ then $L,K$ satisfies the required properties and we can find $m$ as 2. Then we can verify the result.
It is given as a hint to prove by contradiction.
So we can assume that There exist integers $ L,M$ satisfying the conditions but for every integer $m$ with $L<m \leq K$, $ \frac{m}{n}$ cannot be an upper bound for $E$ .That means  there exists $  x \in E $ such that $x >\frac{m}{n}$ .
I can't proceed from this step. It is also given that we have to use induction.
 A: Define the set $ NE = \{n x  \ | \ x \in E\} $. Now clearly $ x \le  \frac {K}{n}$ for every $x \in E$ and hence $ nx \le K  $ whence it follows that $K$ is an upper bound for $ NE $. Then $l =  \sup NE$ exists. 
Now consider the set $A = \{ n \in \Bbb N \ | n \ge l \ \} $. This set is non-empty since it contains $K$. By the Well-Ordering principle this set has a least element and call this, $m$. 
Firstly, 
$$\forall x\in E \;\; m \ge l \ge nx \implies x \le \frac m n \;\; \forall x\in E  $$
Secondly, suppose $ \dfrac{m - 1}{n} $ is an upper bound for $E$. Then, 
$$ \forall x \in E \;\; x \le \dfrac{m - 1}{n} \implies nx \le m - 1  $$
Then $m - 1$ is an upper bound for  $NE$ and hence $l \le m - 1$. But then $m - 1 \in A$ and $m - 1 \lt m$ contradicting the choice of $m$. So there you have it. 
I have not used induction but I have used the equivalent condition the Well-Ordering Principle. 

Note all we have done here is to divide the real line into segments of size $\frac 1 n$. You can obviously do this. This gives an "index" if you will of the real line. Now the supremum of the set $E$ exists and is on the real line. But this must lie between two indices. That is $ \dfrac{m - 1}{n} \lt l \le \dfrac{m }{n}$. 
