(We are proving the claim for $n \geq 1$, and we have $F_1 = 1, F_2 = 1$ and $F_n = F_{n-1} + F_{n-2}$.)
The proof proceeds by induction on $n$.
Base Case: We have that $F_1 =1$ and $F_2 = 1$, and both clearly only have a factor of $1$. So, this case holds.
Inductive Step: We assume the claim holds for $n=k$ and show this implies the claim holds for $n=k+1$.
For the sake of contradiction, let's assume $F_{k+1}$ and $F_{k+2}$ are not relatively prime. Then, there is some integer $d >1$ such that $d|F_{k+1}$ and $d|F_{k+2}$.
We have that $F_{k+2} = F_{k+1} + F_k$. It follows that since $d|F_{k+2}$ and $d|F_{k+1}$, we must also have $d|F_k$. This implies that $F_{k+1}$ is not relatively prime with $F_k$, since both share a factor $d > 1$. This contradicts the inductive hypothesis, so we conclude $F_{k+1}$ and $F_{k+2}$ must be relatively prime.
You can also use the Bezout identity in your induction to have a direct proof.
$\gcd(F_{n+1},F_n)=1\iff \exists (a,b)\mid aF_{n+1}+bF_n=1$
Then $F_{n+2}=F_{n+1}+F_n$ after multiplying by $b$ we get
$\implies bF_{n+2}=bF_{n+1}+1-aF_{n+1}\\\implies bF_{n+2}+(a-b)F_{n+1}=1\\\implies \gcd(F_{n+2},F_{n+1})=1$