\documentclass[a4paper,11pt]{book}
\usepackage[polish]{babel}
\usepackage[utf8]{inputenc} 
\usepackage[T1]{fontenc}
\usepackage{fourier}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{geometry}

\geometry{width=126mm}



\begin{document}

\begin{proof}
Niech $\mathbf{v} = [x,y,z]$. Z założenia 
$\mathbf{u} = \alpha\mathbf{v}$ wynika, że 
$\mathbf{u} = [\alpha x,\alpha y,\alpha z]$.\\
Obliczamy długość wektora $\mathbf{u}$:
\begin{equation}
\begin{split}
|\mathbf{u}| &= \sqrt{(\alpha x)^2 + (\alpha y)^2 + (\alpha z)^2} = 
\sqrt{\alpha^2(x^2 + y^2 + z^2)} = \\[1.5mm]
&= |\alpha|\sqrt{x^2 + y^2 + z^2} = |\alpha||\mathbf{v}|
\end{split}
\end{equation}
co należało udowodnić.
\end{proof}

\end{document}