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 $d$ be their gcd.
Then $d$ divides their sum $p$,
so $d$ can be only 1 or $p$.

If $d = p$, then $p$ divides both $a$ and $b$.
Since both of these are positive,
they are each at least $p$,
so their sum is at least $2p$.

I realize that this is a restatement of mixedmath's answer.

This can easily be generalized to show that
this holds for the sum of $k$ positive integers
for $k \ge 2$.
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, $d=1$.