I would like to have feedback on the overall quality of the following proof.
Question: Prove that $x^5+7x-1=0$ has a unique solution in $[0,1]$.
Proof: Let $f(x)=\frac{1-x^5}{7}$ and note that any solution of $f(x)=x$ is in fact a solution of our polynomial equation. Since $[0,1]$ is closed in $\mathbb{R}$, which is complete, it follows that $[0,1]$ is also complete. Since our polynomial equation is algebraic we use the Euclidean metric, denoted by $d$. Together, $([0,1],d)$ is a complete metric space.
To satisfy Banach's fixed point theorem we require that $f$ is a contraction on $[0,1]$:
$d(f(x),f(y))=|f(x)-f(y)|=\frac{1}{7}|x^5-y^5|=\frac{1}{7}|x-y|\cdot|x^4+x^3y+x^2y^2+xy^3+y^4|\leq \frac{5}{7}|x-y|=\frac{5}{7}d(x,y)$
for all $x,y\in[0,1]$. Thus $f$ is a contraction on $[0,1]$. Therefore, by Banach's fixed point theorem, $f$ must have a unique fixed point in $[0,1]$ and so the polynomial equation must have a unique solution in $[0,1]$.
All suggestions for improvements are welcome!
A hostile reviewer could raise the following issues:
The only really serious issue is point $4$. The function $g : [0,1] \rightarrow \mathbb{R}$ given by $g(x) =2$ is a contraction, but it has no fixed points. Nevertheless, it would serve you well to eliminate the remaining issues.
It is not necessary to apply this level of formalism in your daily work, but apply it when you need to remove any doubt about your ability to justify your reasoning.