$0$ is obtained via the empty set.
We'll proceed by "simultaneous induction" on the positive and negative integers.
To build up positive base cases we note that $$1=(-2)^0\quad \quad 2=(-2)^2+(-2)^1\quad \quad 3= (-2)^2+(-2)^1+(-2)^0$$
To build up negative base cases we note that $$-1=(-2)^1+(-2)^0\quad \quad -2=(-2)^1\quad \quad -3=(-2)^3+(-2)^2+(-2)^0$$
Now the induction statement we want is "Given that the claim is true for all integers $k$ with $|k|≤n-1$ prove that it is also true for $k=\pm n$."
That plus the base cases will certainly suffice.
To prove the statement, we first note that (using the base cases) we can assume that $n≥4$. Now we distinguish between the cases $n$ even or $n$ odd.
If $n$ is even then $\frac n{-2}$ is an integer with absolute value $<n$ so we can write $$\frac n{-2}=\sum_{i=1}^m(-2)^{a_i}\implies n=\sum_{i=1}^m(-2)^{a_i+1}$$
(here, of course, we are using a proper representation of the smaller number. Thus the $\{a_i\}$ are distinct. If that is the case, then of course the numbers $\{a_i+1\}$ are also all distinct.)
If $n$ is odd then $n-1$ is even and, as before we can write $$\frac {n-1}{-2}=\sum_{i=1}^m(-2)^{a_i}\implies n=\sum_{i=1}^m(-2)^{a_i+1}+(-2)^0$$ and we are done.
The case of $-n$ is more or less identical.
Note that this method is "constructive" in the sense that you can use it to construct the representation of some number, given that you have already got the representations of smaller numbers.
One should be more precise. The statement does not follow merely from the fact that there are infinitely many integers. We shall use some properties of the order relation on the set of real numbers. These properties are immediate from the construction of $\mathbb{R}$. Let $x\in \mathbb{R}$
Existence:
There exists a natural number $n\in \mathbb{N}$ such that $-n\leqslant x < n$ ($\mathbb{R}$ has the Archimedean property). Observe that $\{u\in \mathbb{R}|-n\leqslant u <n\}=\bigcup_{i=1}^{2n} \{v\in\mathbb{R}|i-n-1\leqslant v <i-n\}$. Hence, $x\in \{v\in\mathbb{R}|i-n-1\leqslant v <i-n\}$ for some $i\in \{1,2,\ldots,2n\}$. For such $i$, $\lfloor x \rfloor=i-n-1$
Uniqueness:
Suppose that $k,r \in \mathbb{Z}$ such that $k\leqslant x <k+1$ and $r\leqslant x <r+1$. If $k<r$, then $k+1 \leqslant r$ and we have that $k\leqslant x<k+1\leqslant r \leqslant x<r+1$. This is a contradiction (Because we get x<x). Since exactly one of the statements $k<r$, $k=r$ and $k>r$ is true, WLOG, we conclude that $k=r$.
In essence, the given definition is fine but it can be "simplified" this way: $(k\in\mathbb{Z},\;\; (k\leqslant x<k+1)) \Leftrightarrow (\lfloor x \rfloor = k)$ for all $x\in\mathbb{R}$.
Best Answer
Assume that the equation is $y=mx+b$, where $m,b\in\mathbb{R}$. If either or both of $m$ and $b$ are irrational, then $mx+b$ would be irrational. Hence, the equation cannot have an integer pair as a solution.
Thus, we assume that $m,b\in\mathbb{Q}$. As a consequence, we have $m=\frac{m_1}{m_2}$ and $b=\frac{b_1}{b_2}$, where $m_1,m_2,b_1,b_2\in\mathbb{Z}$ and $m_2,b_2\neq 0$. Substituting these in the equation, we get $y=\frac{m_1}{m_2}x+\frac{b_1}{b_2}$. This equation can be converted into an equivalent equation of the form $px+qr=r$, where $p,q,r\in\mathbb{Z}$. This is called a 'Linear Diophantine equation'.
A well-known result is that any equation of the above form has an integer solution if and only if $GCD(p,q)|r$ (i.e., the greatest common divisor of $p$ and $q$ divides $r$).
Once an equation satisfies the above condition, we can determine an integer solution. This can be done using the 'Extended Euclidean algorithm', which gives a solution to the equation $px+qy=GCD(p,q)$. Let $(x',y')$ be a solution. Since it satisfies the equation, we have $px'+qy'=GCD(p,q)$. Multiply the whole equation by $c=\frac{r}{GCD(p,q)}$. This gives $cpx'+cqx'=c\cdot GCD(p,q)$ or equivalently, $p(cx')+q(cy')=r$. Thus, $(cx',cy')$ is a solution to $px+qy=r$.