Proof of a Stronger Version of Dirichlet's Approximation Theorem If $\alpha$ is a real number and $n$ is a positive integer, there are integers $a$ and $b$ such that $1\leq a \leq n$ and $|\alpha a - b| < \frac{1}{n+1}$.
Here is an attempt of the proof. I'm more or less skeptical of the second case and on whether or not the cases are exhaustive. Ultimately, I am wondering if the proof is alright.
Notation
Let $x \in \mathbb{Z}$. Then I use $[x]$ to denote the floor of $x$. 
A real number $\rho$ can be expressed as the sum of its integer $[\rho]$ and fractional parts $\{\rho\}$ as $\rho = [\rho] + \{\rho\}$ where $0 \leq \{\rho\} < 1$.
Proof
Consider the $n+2$ fractional parts $\{j\alpha\}$ where $j = 0,\cdots,n+1$. 
Then partition the interval $[0,1)$ into the $n+1$ subintervals 
$$\Big[\frac{k-1}{n+1},\frac{k}{n+1}\Big)$$
Then each fractional part lies in one of the subintervals.
Since there are $n+2$ numbers but only $n+1$ intervals, the Pigeonhole principle says that at least two of these numbers lie in the same interval. Further, because each interval has length $\frac{1}{n+1}$ and does not include its right endpoint, each point in a given interval will be less than $\frac{1}{n+1}$.
Hence, there are integers $0 \leq p < q \leq n+1$ such that $|\{q\alpha\} - \{p\alpha\}| < \frac{1}{n+1}$. 
Upon manipulating the inequality of $p$ and $q$ we get 
$$0 \leq p \leq q-1 \leq n$$
First, assume that $p< q-1$. Then $1 \leq q-1-p\leq n$. 
Therefore, let $a = q-1-p$ and $b = [q\alpha]-[p\alpha]-\alpha$. 
Now, we have
$$\begin{align}|a\alpha-b| &= |(q-1-p)\alpha - ([q\alpha]-[p\alpha]-\alpha)| \\\\ &= |(\alpha q - [\alpha q]) - (\alpha p - [\alpha p])| \\\\ &= |\{\alpha q\} - \{\alpha p\}| \\\\ &< \frac{1}{n+1}\end{align}$$
So the given choices for $a$ and $b$ suffice.
Now suppose that $q-1 = p$. Then $1 = q-p$, which satisfies the inequality $1 \leq q-p \leq n$. Hence, let $a = q-p$ and $b=[\alpha q] - [\alpha p]$. Then
$$\begin{align}|a \alpha - b| &= |(q-p)\alpha - ([\alpha q] - [\alpha p])|  \\\\ &= |(\alpha q-[\alpha q]) - (\alpha p - [\alpha p])| \\\\ &= |\{\alpha q\} - \{\alpha p\}| \\\\ &< \frac{1}{n+1}\end{align}$$
Thus the given choies for $a$ and $b$ suffice.
Upon exhausting all possibilities, we conclude that one of  first $n$ multiples of a real number $\alpha$ must be within $\frac{1}{n+1}$ of an integer.
 A: My comment is wrong: the problem in your proof is that $b$ must be an integer, and in your first case, where $p < q-1$, you have $b$ to be $([q\alpha] - [p\alpha] - \alpha)$, but since $\alpha$ is a real number, b is not necessarily an integer. 
Here is how I did the proof (it's really similar and only differs in one small aspect):
Consider the $n+2$ numbers: $0, \{\alpha\}, \{2\alpha\}, ..., \{n\alpha\}, 1$. 
Now divide the closed interval $[0,1]$ into $n+1$ partitions:
$[0, \frac{1}{n+1}), [\frac{1}{n+1}, \frac{2}{n+1}), ..., [\frac{n}{n+1}, 1]$
By the pigeonhole principle, since all $n+2$ numbers fall in the interval $[0,1]$, there must be one of the $n+1$ subintervals that contains two numbers. Now, there are two cases:

Case 1: The two numbers are in the first $n$ subintervals, subinterval $i$, $[\frac{i-1}{n+1},\frac{i}{n+1})$.
Let $p, q \in \mathbb{Z}$ such that $0\leq p \lt q \leq n$ without loss of generality. We have $|\{q\alpha\} - \{p\alpha\}| \lt \frac{1}{n+1}$
Thus, by definition of $\{\}$, we have $|q\alpha - \lfloor q\alpha \rfloor - (p\alpha - \lfloor p\alpha \rfloor)| \lt \frac{1}{n+1}$
Rearranging, we have $|(q-p)\alpha - (\lfloor q\alpha \rfloor - \lfloor p\alpha \rfloor)| \lt \frac{1}{n+1}$
Take $a = q-p$ and $b = \lfloor q\alpha \rfloor - \lfloor p\alpha \rfloor$, both of which are integers, to satisfy the inequality.
Case 2: The two numbers are in the last subinterval, $[\frac{n}{n+1}, 1]$. One of the numbers must be 1. Let the other number be $\{p\alpha\}$, for $p \in [0,n]$. 
Thus, we have $|\{p\alpha\} - 1| \leq \frac{1}{n+1}$. Note that it is less than or equal to because this is the only closed interval. 
Simplifying as we did above, we get $|p\alpha - (\lfloor p\alpha \rfloor + 1)| \leq \frac{1}{n+1}$. 
Take $a = p$ and $b = \lfloor p\alpha \rfloor + 1$, satisfying the constraints. QED. 
A: Your proof is wrong particularly in the first case:

First, assume that $p<q−1$. Then $1 \le q−1−p \le n$.

This is clearly wrong, take $p=1$, $q=2$. Then, $q-1-p=0 < 1$.
You can intuitively understand why the proof is wrong by noticing that you can have $a>n$ when $p=0$ and $q=n+1$, which would invalidate the entire proof.
