I have seen this being used multiple times in some proofs, but I am not able to prove it and I can't find a proof of this statement.
So basically I'm asking if anyone got a source proving this (or even more general) statement.
Edit: $d$ should be odd and greater than $5$.
On
"single real root close to $1$" might mean that there is only one real root, which is in fact close to $1$, or that there is only one real root in some neighborhood of $1$, possibly other real roots elsewhere.
For the second: You could use Rouche's theorem to show that if $\epsilon$ is small enough then there is exactly one complex root in $D(1,r)$. Since $P(\overline z)=\overline{P(z)}$ that root must be real.
(For the first: More or less the same argument shows that if $d$ is even and $\epsilon$ is small enough there are at least two real roots (one close to $-1$.)
For any polynomial Q a number $x_0$ is a simple root if and only if $Q(x_0)=0$ and $Q'(x_0)\neq 0$.
Also recall that the zeros of a complex polynomial depend continuously on the coefficients.
Now let $P_\epsilon(x)=x^d-1 +\epsilon p(x)$, where $p$ is another polynomial. For $\epsilon=0$ this has 1 as a root. Thus, for small $\epsilon$ $P_\epsilon$ has a root $x_\epsilon$ close to 1. Moreover, \begin{equation} P_\epsilon'(x_\epsilon)=dx_\epsilon^{d-1}-\epsilon p'(x_\epsilon), \end{equation} which for $\epsilon$ close to $0$ is close to $d$ and hence not $0$. Therefore, for $\epsilon$ close to $0$, $P_\epsilon$ has a simple root close to 1.