My defintion of fixed point theorem (FPT)
If there exists an interval $[a, b]$ such that $\forall x \in [a, b], \: g(x) \in [a, b]$ AND $\forall x \in [a, b], \: g'(x) \leq L \lt 1$, then $g(x)$ has a unique fixed point $\zeta \in [a, b]$.
Assume the conditions and result of FPT for a function $g(x)$ with interval $[a, b]$ (i.e., for any $x_0 \in [a, b]$ and iterating $x_{k+1} = g(x_k)$ for $k = 0, 1, \: ...$ we get $lim_{k \to \infty} x_k = \zeta$)
Let $g(c) = c'$ for some $c \in [a, b]$, then we have $$ \begin{alignat}{2} c' - x_{k+1} &= g(c) - g(x_k)\\ &= g(\alpha_k)(c - x_k) \quad \text{By MVT for some} \: \alpha_k \in [x_k, c] \end{alignat} $$
Since $\forall x \in [a, b], \: g'(x) \leq L \lt 1$, it follows that $$ \begin{alignat}{2} 0 \leq |c' - x_{k+1}| &\leq L|c - x_k|\\ &\leq L^2|\beta_{k-1} - x_{k-1}|\\ &\leq L^3|\beta_{k-2} - x_{k-2}|\\ &\leq \cdots\\ &\leq L^{k+1}|\beta_0 - x_0|\\ \end{alignat} $$
Where each $\beta_{i+1} = g(\beta_i)$ for some $\beta_i \in [a, b]$ and $g(\beta_{k-1}) = c$ for some $\beta_{k-1} \in [a, b]$. With squeeze theorem, we have $$ \begin{alignat}{2} &\lim_{k \rightarrow \infty} |c' - x_{k+1}| = 0\\ \Longleftrightarrow \quad &\lim_{k \rightarrow \infty} |c'| = \lim_{k \rightarrow \infty} |x_{k+1}|\\ \Longleftrightarrow \quad &|c'| = \zeta \end{alignat} $$
This does not make sense since it means we can choose any $c' \in [a, b]$ and it will be the unique fixed point $\zeta \in [a,b]$. Where does this proof go wrong?