The exercise is to prove that $A \ne 0 \land A \times B = A \times C \land A \cdot B = A \cdot C \implies B = C$.
Let's assume $B \ne C$. Then we know that $\tag{1} \|B\| \ne \|C\|$
On the other hand, by the hypothesis and Lagrange's identity we have
$$\|A \times B \|^2 = \|A\|^2\|B\|^2 - (A \cdot B)^2=\|A\|^2\|C\|^2 - (A \cdot C)^2 = \|A \times C \|^2$$ $$\implies \|A\|^2\|B\|^2=\|A\|^2\|C\|^2$$ $$\implies \|B\|^2=\|C\|^2$$ $$\implies \|B\|=\|C\|$$
The first implication is because $A \cdot B = A \cdot C$, and the second one because $A \ne 0$. The third implication is because the norm is not negative. That finally contradicts with $(1)$, and I would call the proof complete.
Is that correct? I'm quite insecure with my proofs by contradiction, and I'm not sure if there is a way to make sure I can validate them myself in general?
Edit:
I think my main doubt comes from a possibility that the hypothesis is false in itself, so whatever we add to it would also make it an absurd, so we can simply conclude that $\bot \implies \top$.