Fake Proofs in Mathematics

Suppose $$u_1+u_2+u_3=0,\qquad u_i\in U_i.$$ Rearranging gives $u_1=-(u_2+u_3)$, so $u_1\in U_1\cap(U_2+U_3)$. Since the pairwise intersections are zero, $$\begin{aligned} U_1\cap(U_2+U_3) &=(U_1\cap U_2)+(U_1\cap U_3)\\ &=\{0\}+\{0\}=\{0\}. \end{aligned}$$ Therefore $u_1=0$. The relation becomes $u_2=-u_3$, so $u_2\in U_2\cap U_3=\{0\}$, and hence $u_2=u_3=0$. The only relation is the trivial one, so the sum is direct. $\square$