Consecutive Numbers are coprime

Your proof looks good. Using the method of contradiction is not a bad idea here but you could have skipped that in your prove.

Given that $n$ and n+1 are two consecutive integers. Now suppose $gcd(n,n+1)=p$. Then p|n and $p|n+1$. Which implies that $p|n+1-n$ or $p|1$. There is no number which divides 1 except 1. So $p=1$ or you can say $gcd(n,n+1)=p=1$. Which implies $n$ and $n+1$ are coprime.

Notice that I have not used contradiction anywhere.

Your proof is correct, but there is a simpler one, that doesn't work by contradiction and doesn't need greatest common divisors explicitly.

If a prime $p$ divides $n$ then dividing $n+1$ by $p$ leaves a remainder of $1$, so $p$ does not divide $n+1$. That means the factorizations of $n$ and $n+1$ have no prime in common, so $n$ and $n+1$ are relatively prime.

But ... this proof does use unique factorization, which is usually proved by thinking about greatest common divisors. So they are there, behind the scene.

We $ $ have $\,n\mid \color{#c00}{\overbrace{n\!+\!1}^{\large m}}-1.\ $ Your (correct) proof easily generalizes as follows.

Generally $\,n\mid \,k\:\color{#c00}m-1\,\Rightarrow\, k\,m-j\,n = 1,\ $ so $\,\ d\mid m,n\,\Rightarrow\, d\mid 1,\ $ so $\ \gcd(m,n) = 1$

$ $ i.e. $\ {\rm mod}\,\ n\!:\,\ m^{-1}\,$ exists $\,\Rightarrow\, \gcd(m,n) = 1.\,$ The converse is also true (Bezout gcd identity)

Remark $\ $ We can also use (a single step of) the Euclidean algorithm

$$ \gcd(km,n) = \gcd(jn\!+\!1,n) = \gcd(1,n) = 1\,\Rightarrow\, \gcd(m,n) = 1$$