How do you prove this if x is a block vector with two vector elements, $\begin{bmatrix} a \\ b \end{bmatrix}$ where a and b are vectors of size n and m respectively?
\begin{align} \| x\|= \sqrt{\| a \|^2 + \| b\|^2 } = \left\|\begin{bmatrix} \| a\| \\ \| b\| \end{bmatrix}\right\| \end{align}
The norm $||\boldsymbol{x}||$ of a vector $\boldsymbol{x}=[\boldsymbol{x}_1,\boldsymbol{x}_2]^T=[\alpha_1,...\alpha_N,\beta_1,...,\beta_M]^T$ is given by
$$||\boldsymbol{x}||=\sqrt{\boldsymbol{x}^T\boldsymbol{x}}=\sqrt{\alpha^2_1+...+\alpha_N^2+\beta_1^2+...+\beta_M^2}=\sqrt{\boldsymbol{x}^T_1\boldsymbol{x}_1+\boldsymbol{x}^T_2\boldsymbol{x}_2}=\sqrt{||\boldsymbol{x}_1||^2+||\boldsymbol{x}_2||^2}.$$
Now, for the second part start with the right-hand side.
$$||[||\boldsymbol{x}_1||,||\boldsymbol{x}_2||]^T||=||[\sqrt{\alpha^2_1+...+\alpha_N^2},\sqrt{\beta_1^2+...+\beta_M^2}]||$$ $$=\sqrt{\sqrt{\alpha^2_1+...+\alpha_N^2}^2+\sqrt{\beta_1^2+...+\beta_M^2}^2}$$ $$=\sqrt{\alpha^2_1+...+\alpha_N^2\beta_1^2+...+\beta_M^2}=||\boldsymbol{x}||$$