Let $S$ be a subspace of a normed linear space, $X$ and $x_0\in X\backslash S$. Consider the subspace spanned by $M,$ i.e. \begin{align} M:=[S\cup \{x_0\}]=\{m=x+\alpha\,x_0:\,x\in S,\;\text{for some}\;\alpha\in\Bbb{R}\} \end{align} I want to show that $m$ is unique.
MY TRIAL
Let $m\in M,$ then there exists $\alpha\in\Bbb{R} $ such that $m=x+\alpha\,x_0.$ Suppose we have another representation, then there exists $\beta\in \Bbb{R},\;\beta\neq \alpha,$ such that $m=x+\beta\,x_0.$ Thus, \begin{align} x_0=0 \;\text{which implies}\;x_0\in S,\;\text{contradiction, since }\; x_0\notin S.\end{align} Please, I'm I right? If not, could you please, provide an alternative proof?
Credits to @Theo Bendit for the hints.
Corrected:
Let $x_0\in X\backslash S$ be arbitrary. Suppose that $m\in M, $ then there exists $x\in S\;\text{and}\;\alpha\in\Bbb{R} $ such that $m=x+\alpha\,x_0.$ Assume there is another representation, then there exists $y\in S\;\text{and}\;\beta\in \Bbb{R}$ such that $m=y+\beta\,x_0,$ where $x\neq y$ or $\alpha\neq\beta$. If $x\neq y$, then $\beta \neq \alpha.$ Otherwise, if $\beta \neq \alpha$ then $x=y$ or $x\neq y.$ In both cases, $\beta \neq \alpha$. Thus, \begin{align} \left(x+\alpha\,x_0=y+\beta\,x_0\right)&\implies (x-y)=(\beta-\alpha)\,x_0\\&\implies x_0=\dfrac{1}{\beta-\alpha}(x-y)\in S,\;\text{where}\;\beta-\alpha\neq 0\;\text{and}\;x-y\in S,\\&\implies x_0\in S,\;\text{since }\;S\;\text{is a subspace of a normed linear space }\\&\implies\text{a contradiction }\end{align} Hence, the representation $m=x+\alpha x_0,\;$ for $\;m\in M$ is unique.