Uniqueness of minimal norm solution. $\newcommand{\R}{\operatorname{Ran}} \newcommand{\K}{\operatorname{Ker}}\newcommand{\b}{\mathbf}$
Let an equation $A\b x = \b b$ has a solution, and let $A$ has non-trivial kernel. 
Prove that there exist a unique solution $\b x_0$ of $A\b x =\b b$ minimizing the norm $ ||\b x||$.

Any solution to  $A \b x = \b b$ can be written as $\b x_1 - \b x_h$ for $\b x_1$ being a solution of $A\b x = \b b$  and $\b x_h \in \K A$. So I need to minimize $||\b x_1 -\b x_h||$ for varying $\b x_h$. 
Since $||\b x_1 -\b x_h||$ is distance from the subspace $\K A$ to $\b x_1$, therefore it will be minimum when $\b x_h = P_{\K A} \b x_1$. So the minimum occurs at $\b x_0 =\b x_1 - P_{\K A} \b x_1 = P_{(\K A)^\perp} \b x_1$, for any $x_1 : A\b x_1 = \b b$. 
How can I prove the uniqueness of this solution ?

If there are any distinct and parallel solutions (regardless of norm), then $\mathbf b=0$ (*), and $0$ is the unique minimal norm solution.
If there are two distinct non-parallel solutions with a given norm, then their average has smaller norm (the triangle inequality is strict for non-parallel vectors). So if there is a solution with minimal norm, it must be unique.
Proof of (*): Say $A\mathbf x_1=A(\lambda\mathbf x_1)=\mathbf b$ and $\lambda\neq1$. Then $$\mathbf b=A(\lambda\mathbf x_1)=\lambda A\mathbf x_1=\lambda \mathbf b$$
