Let $X$ be a normed linear space. Suppose that
$(a)$ $M$ is a closed subspace of $X$;
$(b)$ $x_0\in X\setminus M$, and
$(d)$ $d=\text{dist}(X,M)=\inf\{\rVert x_0-m\lVert\;m\in M\}$
We define $M_1=\text{span}\{M, x_0\}$, then each $x\in M_1$ can be written as $x=m_x+t_xx_0$ for some $m_x\in M$ and $t_x\in\mathbb{R}(\text{or}\;\mathbb{C})$.
I must prove that this representation is unique. Suppose that $$m_x+t_x x_0=n_x+s_x x_0\quad(m_x\ne n_x)\quad s_x\in\mathbb{R}\;\text{or}\;\mathbb{C},$$ then we have $$m_x-n_x=x_0(s_x-t_x)\in M$$ but then $$\frac{1}{s_x-t_x}(s_x-t_x)x_0=x_0\in M$$ Absurd! It's correct?
Second proof
We suppose that $$m_x+t_xx_0=n_x+s_x x_0\quad\text{where}\quad m_x,n_x\in M, t_x,s_x\in \mathbb{R}\;\text{or}\;\mathbb{C}$$ then,
$$(m_x-n_x)+(t_x-s_x)x_0=0\iff m_x=n_x\;\text{and}\quad(t_x-s_x)x_0=0.$$
Now $(t_x-s_x)x_0=0$ iff $t_x=s_x$ or $x_0=0$, but if $x_0=0$ then $x_0\in M$, absurd. Correct?
All the logical arguments follows, however the thesis is not being properly negated.
Note that, to show that the representation is unique, is to show that both $m_x = n_x$ and $s_x = t_x$. It happens that the the negation of this statement is $m_x \neq n_x$ or $s_x \neq t_x$, and you are just assuming that $m_x \neq n_x$.