Question: I feel this is an intuitive result.
If, for example, I was working with the prime number $11$, I could split it in the following ways: $\{1, 10\}$, $\{2, 9\}$, $\{3, 8\}$, $\{4, 7\}$, $\{5, 6\}$.

Then clearly there is no way that the $2$ numbers can have a $\gcd$ of anything other than $1$. However, I am sort of lost on how to start a formal proof for this. Any pointers would be much appreciated.

Answer: 
Let $c$ be the gcd. Then $c$ divides $a$ and $b$, hence it divides $a+b$, a prime number.
Suppose $a,b$ are positive integers whose sum is a prime, $p$. Then $a+b = p$. Also, because $a, b$ are both positive integers that add up to $p$, then $a,b < p$. This implies that $\gcd(a,p) = 1$ and $\gcd(b,p) = 1$. Thus we can write 1 as a linear combination of both $a$ and $p$ or $b$ and $p$. Let us do it the first way. $ax + py = 1$ for some $x, y \in \mathbb{Z}$. From our original assumption, $a+b = p$. Thus we can substitute. $ax + (a+b)y = 1$. This implies $ax + ay +by = 1$. Grouping like terms we get $a(x+y) + b(y) = 1$. Thus 1 can be written as a linear combination of $a$ and $b$. This implies that $\gcd(a,b) = 1$. We have shown what we wanted to show.