Let $A=\matrixcolumns{A}{n}$. $\linearsystem{A}{\vect{b}}$ is consistent, so we know the system has at least one solution (Definition CS). We would like to show that there are no more than one solution to the system. Employing Proof Technique U, suppose that $\vect{x}$ and $\vect{y}$ are two solution vectors for $\linearsystem{A}{\vect{b}}$. By Theorem SLSLC we know we can write,
\begin{align*}
\vect{b}&=
\vectorentry{\vect{x}}{1}A_1+
\vectorentry{\vect{x}}{2}A_2+
\vectorentry{\vect{x}}{3}A_3+
\cdots+
\vectorentry{\vect{x}}{n}A_n\\
\vect{b}&=
\vectorentry{\vect{y}}{1}A_1+
\vectorentry{\vect{y}}{2}A_2+
\vectorentry{\vect{y}}{3}A_3+
\cdots+
\vectorentry{\vect{y}}{n}A_n
\end{align*}
Then
\begin{align*}
\zerovector
&=\vect{b}-\vect{b}\\
&=
\left(
\vectorentry{\vect{x}}{1}A_1+
\vectorentry{\vect{x}}{2}A_2+
\cdots+
\vectorentry{\vect{x}}{n}A_n
\right)
-
\left(
\vectorentry{\vect{y}}{1}A_1+
\vectorentry{\vect{y}}{2}A_2+
\cdots+
\vectorentry{\vect{y}}{n}A_n
\right)\\
&=
\left(\vectorentry{\vect{x}}{1}-\vectorentry{\vect{y}}{1}\right)A_1+
\left(\vectorentry{\vect{x}}{2}-\vectorentry{\vect{y}}{2}\right)A_2+
\cdots+
\left(\vectorentry{\vect{x}}{n}-\vectorentry{\vect{y}}{n}\right)A_n
\end{align*}
This is a relation of linear dependence (Definition RLDCV) on a linearly independent set (the columns of $A$). So the scalars must all be zero,
\begin{align*}
\vectorentry{\vect{x}}{1}-\vectorentry{\vect{y}}{1}&=0
&
\vectorentry{\vect{x}}{2}-\vectorentry{\vect{y}}{2}&=0
&
\dots&
&
\vectorentry{\vect{x}}{n}-\vectorentry{\vect{y}}{n}&=0
\end{align*}
Rearranging these equations yields the statement that $\vectorentry{\vect{x}}{i}=\vectorentry{\vect{y}}{i}$, for $1\leq i\leq n$. However, this is exactly how we define vector equality (Definition CVE), so $\vect{x}=\vect{y}$ and the system has only one solution.