Without using dimensionality arguments:
Suppose $(a,b)$ and $(c,d)$ are both orthogonal to $(e,f)\ne (0,0)$.
Then, from the definition of orthogonality: $$ ae+bf =0 $$ and $$ ce+df =0 .$$
If $e=0$, we must have $f\ne 0$, which implies $b=d=0$. Thus, $(a,b)=(a,0)$ and $(c,d)=(c,0)$ are scalar multiples of each other.
If $e\ne 0$, then, from the above system
$$
a=-\textstyle{ f\over e}\, b \quad \text{and}\quad c=-{f\over e}\,d;
$$
whence,
$$(a,b)=(\textstyle{- f\over e} \thinspace b, b)=b(\textstyle{-f\over e}, 1)$$and$$(c,d)=(\textstyle{- f\over e} \thinspace d, d)=d(\textstyle{-f\over e},1).$$
This implies that $(a,b)$ and $(c,d)$ are scalar multiples of each other.
To shorten the proof, we may write as suggested by André Nicolas,
Proof. Let $(V,+,\cdot)_F$ be a vector space over the field $F$. We wish to show that $\forall v\in V$ one has $0\cdot v=\mathbf{0}$, where $0$ is the zero scalar and $\mathbf{0}$ is the zero vector. Let $v$ be an element of the vector space $V$;
By one of the axioms of field addition, $$0\cdot v=(0+0)\cdot v.$$
Since scalar multiplication is distributive over addition, $$(0+0)\cdot v=0\cdot v+0\cdot v.$$
From the previous two equalities we conclude that $$0\cdot v=0\cdot v+0\cdot v.$$
Adding to both sides the inverse element for addition of $0\cdot v$, which we'll denote by $-0\cdot v$: $$0\cdot v+(-0\cdot v)=0\cdot v+0\cdot v+(-0\cdot v).$$
By the inverse axiom, $$\mathbf{0}=0\cdot v+\mathbf{0},$$
hence by the identity axiom, $$0\cdot v=\mathbf{0}.\tag*{$\square$}$$
Best Answer
Let $x\in V$ with $x\ne 0$, and let $a,b\in F$ with $a\ne b$.
Suppose $ax=bx$. \begin{align*} \text{Then}\;\;&ax=bx\\[4pt] \implies\;&ax-bx=0\\[4pt] \implies\;&(a-b)x=0\\[4pt] \implies\;&cx=0,\;\text{where}\;c=a-b\\[4pt] \implies\;&c^{-1}cx=c^{-1}0\qquad\text{[$\,c^{-1}\;$exists since$\;a-b\ne 0\,$]} \\[4pt] \implies\;&x=0\\[4pt] \end{align*} contradiction.
Therefore $ax\ne bx$.