= Solution
Use generators $x,D$ with $Dx-xD=1$. The ordered <monomials> $x^iD^j$ form a <basis>: construct the <Weyl algebra> as the differential <Ore extension> of $\mathbb C[x]$ with $Df=fD+f'$, which has unique ordered polynomial expressions. Giving both generators degree one produces the <Bernstein filtration>, whose <associated graded ring> is $\mathbb C[x,\xi]$. If $a,b\ne0$, their nonzero leading symbols have nonzero product in this polynomial <integral domain>. Thus $ab$ has nonzero leading symbol and cannot vanish. Hence \b[$A_1(\mathbb C)$ has no nontrivial <zero divisors>].
Here is the left-handed <Ore theorem>. Let $S$ be a multiplicative set of <regular elements of a ring>, containing $1$ and excluding $0$. There is an injective <Ore localization> whose elements are left fractions $s^{-1}a$ if and only if
$$
\boxed{\text{for every }a\in A,s\in S,\text{ there are }t\in S,b\in A\text{ with }ta=bs.}
$$
This is the left <Ore condition>. For a <noncommutative domain>, taking $S=A\setminus\{0\}$ gives a <division ring> exactly when this condition holds. With zero-divisor denominators a further left reversibility condition is needed; the regular-denominator statement is the one used here.
Necessity follows by writing $as^{-1}=t^{-1}b$ in the quotient and multiplying by $t$ and $s$, giving $ta=bs$ in the embedded original <ring>. For a construction proving sufficiency, use the <left Ore fraction construction>: pairs $(s,a)$ represent $s^{-1}a$, and $(s,a)$ and $(t,b)$ are equivalent when
$$
us=vt=w\in S,\qquad ua=vb
$$
for suitable $u,v\in A$. Reflexivity and symmetry are immediate. For transitivity, combine $us=vt=w$ and $u't=v'q=w'$ by a common left multiple $hw=kw'\in S$. Then $(hv-ku')t=0$, so regularity gives $hv=ku'$, which combines the numerator equalities and proves transitivity.
Addition uses the common denominator $w$, with numerator $ua+vb$. To multiply $s^{-1}a$ by $t^{-1}b$, choose $v\in S,c\in A$ with $va=ct$; set
$$
(s^{-1}a)(t^{-1}b)=(vs)^{-1}cb.
$$
The <Ore condition> supplies a common refinement whenever a representative or an Ore relation is changed, and regularity cancels the shared denominator, proving independence of choices. The same refinements verify the <ring> axioms; the formulas are simply addition and multiplication after clearing left denominators. The map $a\mapsto(1,a)$ is injective, since equivalence to zero would give $ua=0$ with $u\in S$. Each $s\in S$ becomes invertible. A homomorphism inverting $S$ extends uniquely by $(s,a)\mapsto f(s)^{-1}f(a)$, giving the universal property. For all nonzero denominators in a <noncommutative domain>, a nonzero fraction has inverse $(s^{-1}a)^{-1}=a^{-1}s$, so the quotient is a <division ring>. This is the requested proof sketch, with the fraction equivalence and multiplication made explicit.
It remains to check the left <Ore condition> for $A_1(\mathbb C)$. Its graded polynomial <ring> is <Noetherian> by the <Hilbert basis theorem>. Lifting homogeneous generators of a graded <left ideal> and successively cancelling leading symbols proves that the filtered <algebra> is a <left Noetherian ring>: each subtraction lowers the nonnegative degree, so it terminates.
In any left Noetherian <noncommutative domain>, nonzero principal <left ideals> intersect. Otherwise $Ab\cap Aa=0$ with $a,b\ne0$ would make
$$
Ab\oplus Aba\oplus Aba^2\oplus\cdots
$$
an infinite direct sum of nonzero <left ideals>. To verify directness, reduce a finite relation modulo $Aa$ to make its first summand zero, then cancel $a$ on the right and repeat. Its increasing finite partial sums contradict the <ascending chain condition>. A nonzero element of $Aa\cap As$ therefore has the form $ta=bs\ne0$, with $t\ne0$, proving the required condition. Consequently \b[$A_1(\mathbb C)$ has a classical <ring> of left quotients, and it is a division <ring>].
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