I know that this question already has an accepted answer, but let me give an answer from a purist differential forms perspective. Recall that if $V$ is an $n$-dimensional inner product space, then the Hodge star is a linear isomorphism $\ast : \bigwedge^k V \to \bigwedge^{n-k} V$ for each $0 \leq k \leq n$, satisfying the following:
for $v$, $w \in V$, $\langle v,w\rangle \omega = v \wedge \ast w$ for $\omega = \ast 1$ the generator of $\bigwedge^n V$ satisfying $\omega = e_1 \wedge \cdots \wedge e_n$ for any orthonormal basis $\{e_k\}$ of $V$ with the appropriate orientation (e.g., the volume form in $\bigwedge^n (\mathbb{R}^n)^\ast$);
in particular, $\ast \ast = \operatorname{Id}$ when $n$ is odd.
Also, recall that the inner product on $\bigwedge^k V$ is given by
$$
\langle v_1 \wedge \cdots \wedge v_k, w_1 \wedge \cdots \wedge w_k \rangle = \det(\langle v_i,w_j \rangle).
$$
So, suppose that $V$ is $3$-dimensional, in which case we can define the cross product of $a$, $b \in V$ by
$$
a \times b := \ast (a \wedge b).
$$
Let $u$, $v$, $w \in V$. Then for any $x \in V$,
$$
\langle u \times (v \times w), x \rangle \omega = \langle x, u \times (v \times w) \rangle \omega\\
= \langle x, \ast (u \wedge \ast(v \wedge w)) \rangle \omega\\
= x \wedge \ast \ast (u \wedge \ast(v \wedge w))\\
= x \wedge (u \wedge \ast(v \wedge w))\\
= (x \wedge u) \wedge \ast (v \wedge w)\\
= \langle x \wedge u, v \wedge w \rangle \omega\\
= (\langle x,v \rangle\langle u,w \rangle - \langle x,w \rangle\langle u,v \rangle)\omega\\
= \langle \langle u,w \rangle v - \langle u,v \rangle w, x \rangle \omega,
$$
implying that
$$
\langle u \times (v \times w), x \rangle = \langle \langle u,w \rangle v - \langle u,v \rangle w, x \rangle,
$$
and hence, since $x$ was arbitrary, that
$$
u \times (v \times w) = \langle u,w \rangle v - \langle u,v \rangle w,
$$
as required.
If we let $\alpha=\lambda(v\cdot w)$, then
$$(\alpha u+\beta v)\cdot w=\lambda(v\cdot w)(u\cdot w)+\beta(v\cdot w)=0\\\implies \beta=-\lambda(u\cdot w)$$ as required.
$$(u\times v)\times w=\lambda((v\cdot w) u -(u\cdot w) v)\tag 1$$
(by the way, I have accidentally ended up with a different $\lambda$ to the one given in your derivation, since in that, they seem to have suddenly decided to find an expression for $u\times(v\times w)$ instead of $(u\times v)\times w)$ as we have been working with the whole time. I think my one fits in better.)
For the basis steps, we choose a basis of our choice, $$u=u_{1}i,v=v_{1}i+v_{2}j,w=w_{1}i+w_{2}j+w_{3}k$$
Compute the LHS of $(1)$, $$(u_1v_1(i\times i)+u_1v_2(i\times j))\times(w_1i +w_2j+w_3k)=u_1v_2k\times(w_1i+w_2j)\\=u_1v_2w_1j-u_1v_2w_2i$$
Then compute the RHS of $(1)$, $$(v_1w_1+v_2w_2)u_1i-u_1w_1(v_1i+v_2j)=(v_2w_2u_1)i-(u_1w_1v_2)j$$
Thus $\lambda=-1$, and hence $$(u\times v)\times w=(u\cdot w) v-(v\cdot w) u \tag 2$$
Equivalently $$w \times (u \times v) = u(v \cdot w) - v(u \cdot w)\tag3$$and hence $$a \times (b \times c) = b(a \cdot c) - c(a \cdot b)\tag4$$
Best Answer
Two possible flaws, neither of them honkers, in my view:
(1) Your computation of $\vec{B} (\vec{A} \cdot\vec{C})-\vec{C}(\vec{A}\cdot\vec{B})$ must have some errors -- it should not come out to be the zero vector.
(2) You argument that $\vec{A} \times (\vec{B} \times \vec{C}) = \vec{0}$ also has errors. Draw a picture, and I think you'll see why. $\vec{A} \times (\vec{B} \times \vec{C})$ is perpendicular to $\vec{A}$, and it's perpendicular to $\vec{B} \times \vec{C}$. I don't see how you conclude from this that it must be the zero vector.