(⇒) If $S$ is linearly dependent, then there are scalars $\alpha$ and $\beta$, not both zero, such that $\alpha\vect{u}+\beta\vect{v}=\zerovector$. Suppose that $\alpha\neq 0$, the proof proceeds similarly if $\beta\neq 0$. Now,
\begin{align*}
\vect{u}
&=1\vect{u}&&\knowl{./knowls/property.O.knowl}{\text{Property O}}\\
&=\left(\frac{1}{\alpha}\alpha\right)\vect{u}&&\knowl{./knowls/property.MICN.knowl}{\text{Property MICN}}\\
&=\frac{1}{\alpha}\left(\alpha\vect{u}\right)&&\knowl{./knowls/property.SMA.knowl}{\text{Property SMA}}\\
&=\frac{1}{\alpha}\left(\alpha\vect{u}+\zerovector\right)&&\knowl{./knowls/property.Z.knowl}{\text{Property Z}}\\
&=\frac{1}{\alpha}\left(\alpha\vect{u}+\beta\vect{v}-\beta\vect{v}\right)&&\knowl{./knowls/property.AI.knowl}{\text{Property AI}}\\
&=\frac{1}{\alpha}\left(\zerovector-\beta\vect{v}\right)&&\knowl{./knowls/definition.LI.knowl}{\text{Definition LI}}\\
&=\frac{1}{\alpha}\left(-\beta\vect{v}\right)&&\knowl{./knowls/property.Z.knowl}{\text{Property Z}}\\
&=\frac{-\beta}{\alpha}\vect{v}&&\knowl{./knowls/property.SMA.knowl}{\text{Property SMA}}
\end{align*}
which shows that $\vect{u}$ is a scalar multiple of $\vect{v}$.
(⇐) Suppose now that $\vect{u}$ is a scalar multiple of $\vect{v}$. More precisely, suppose there is a scalar $\gamma$ such that $\vect{u}=\gamma\vect{v}$. Then
\begin{align*}
(-1)\vect{u}+\gamma\vect{v}
&=(-1)\vect{u}+\vect{u}\\
&=(-1)\vect{u}+(1)\vect{u}&&\knowl{./knowls/property.O.knowl}{\text{Property O}}\\
&=\left((-1)+1\right)\vect{u}&&\knowl{./knowls/property.DSA.knowl}{\text{Property DSA}}\\
&=0\vect{u}&&\knowl{./knowls/property.AICN.knowl}{\text{Property AICN}}\\
&=\zerovector&&\knowl{./knowls/theorem.ZSSM.knowl}{\text{Theorem ZSSM}}
\end{align*}
This is a relation of linear of linear dependence on $S$ (Definition RLD), which is nontrivial since one of the scalars is $-1$. Therefore $S$ is linearly dependent by Definition LI.
Be careful using this theorem. It is only applicable to sets of two vectors. In particular, linear dependence in a set of three or more vectors can be more complicated than just one vector being a scalar multiple of another.