Proof using pigeonhole and greatest integer (floor) function. The question is to prove that if m is a positive integer then, 
$$[mx] = [x] + \left[x+\frac{1}{m}\right] +\left[x+\frac{2}{m}\right] + \cdots + \left[x+\frac{(m-1)}{m}\right] $$
for $x \in \mathbb{R}$. Where $[x] =n$ such that $ n \leq x <n+1$
I'm given that the solution should use the pigeonhole principle. So I need to look for $n$ boxes where I'm trying to stuff $n+1$ things (my understanding of the pigeonhole principle). If I look at the fractional part of x I see $\{x\} \in [0,1)$ where $\{x\}$ is the fraction part of x. So $m\{x\} \in [0,m)$. I can break this interval into $[0,1) [1,2) \ldots [m-1,m)$ or I can divide everything by m and get $[0,\frac{1}{m}) [\frac{1}{m},\frac{2}{m}) \ldots [\frac{m-1}{m},1)$ and get m intervals of length $\frac{1}{m}$ which is my "n" boxes. 
I am however stuck with regards to the "$n+1$" objects to put in the box and how to relate this observation to the question. Can anybody provide a hint or point out a flaw in what I have so far? 

I wouldn’t use the pigeonhole principle: it’s not needed. You know that there is a unique integer $n$ such that $0\le n<m$ and $\frac{n}m\le\{x\}<\frac{n+1}m$. Now consider $x+\frac{k}m$, where $0\le k<m$:
$$x+\frac{k}m=\lfloor x\rfloor +\{x\}+\frac{k}m\;,$$ so
$$\lfloor x\rfloor+\frac{n+k}m\le x+\frac{k}m<\lfloor x\rfloor +\frac{n+1+k}m\;.\tag{1}$$
What does $(1)$ tell you about $\left\lfloor x+\frac{k}m\right\rfloor$? Specifically, for how many values of $k$ will it be $\lfloor x\rfloor$, and for how many will it be $\lfloor x\rfloor+1$? If you can answer that, you’ll have a good handle on the righthand side of your desired equation.
As for the lefthand side, start from the fact that $\lfloor mx\rfloor=\big\lfloor m(\lfloor x\rfloor+\{x\})\big\rfloor$.
Most of the Missing Detail: If $0\le k\le m-n-1$, $(1)$ tells us that 
$$\lfloor x\rfloor\le\lfloor x\rfloor+\frac{n}m\le \lfloor x\rfloor+\frac{n+k}m\le x+\frac{k}m<\lfloor x\rfloor+\frac{n+k+1}m=\lfloor x\rfloor+1\;;$$
after getting rid of the excess baggage, we have $$\lfloor x\rfloor\le x+\frac{k}m<\lfloor x\rfloor+1\;,$$ and therefore $$\left\lfloor x+\frac{k}m\right\rfloor=\lfloor x\rfloor\;.\tag{2}$$
On the other hand, if $m-n\le k\le m-1$, $(1)$ implies that 
$$\lfloor x\rfloor+1\le\lfloor x\rfloor+\frac{n+k}m\le x+\frac{k}m<\lfloor x\rfloor+\frac{n+1+k}m\le\lfloor x\rfloor+\frac{n+m}m<\lfloor x\rfloor+2$$
and hence that $$\left\lfloor x+\frac{k}m\right\rfloor=\lfloor x\rfloor +1\;.\tag{3}$$
Now $(2)$ holds for $m-n$ values of $k$, and $(3)$ holds for the other $n$ values of $k$, so 
$$\begin{align*}\sum_{k=0}^{m-1}\left\lfloor x+\frac{k}m\right\rfloor&=(m-n)\lfloor x\rfloor+n(\lfloor x\rfloor+1)\\
&=m\lfloor x\rfloor+n\;.
\end{align*}$$
But $$\begin{align*}
\lfloor mx\rfloor&=\big\lfloor(m\lfloor x\rfloor+\{x\})\big\rfloor\\
&=\big\lfloor m\lfloor x\rfloor+m\{x\}\big\rfloor\\
&=m\lfloor x\rfloor+\lfloor m\{x\}\rfloor\;,
\end{align*}$$
since $m\lfloor x\rfloor$ is an integer.
Now you need only show that $\lfloor m\{x\}\rfloor=n$ to finish the argument. Go back to the beginning to recall how $n$ was defined.
