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's show the contrapositive, because why not? 

So we want to show that if $a,b>0$ and $\gcd(a,b) \neq 1$, then their sum is not prime. 

Suppose that $\gcd(a,b) = d > 1$. Then $a = a'd$ and $b = b'd$ for some $a',b'$ natural numbers. But then $a + b = da' + db' = d(a' + b')$, and as each of $d,a',b' \geq 1$, we have that $a + b \geq 2d$, but is divisible by $d$. Thus it is not prime. $\diamondsuit$

Thus if the sum of positive integers is prime, then their gcd is $1$.
Complete short proof:

Let $\gcd(a,b)=d$. Then, $d \mid a$ and $d \mid b$. Thus, $d \mid a+b=p$. Thus, $d=1$ or $d=p$. Since $d \mid a$, $d\le a<a+b=p$. Thus, $d=1$.