In my book I have the implication: If $gcd(a,b)=1$ and $c|(a+b)$, then $gcd(c,a)=gcd(c,b)=1$. It gives me a hint that begins by supposing that $gcd(a,c)=gcd(b,c)=d$.
But in my opinion, I do not think this is valid to assume for two reasons. First, I interpret the question's conclusion, "[...] $gcd(c,a)=gcd(c,b)=1$" as two parts: Establishing that $gcd(a,c)=gcd(b,c)$, and then showing that this equality is $1$. If I am wrong, could someone explain to me why this is valid to assume?
Note: I am not asking anyone to assist me with constructing a proof, just the above concern is what bothers me.
I'm not really sure what the problem you're having is. What you explain as your objection is in fact just the correct way of interpreting the hint. Maybe it will help if we go through it really carefully:
You want to prove that $\operatorname{gcd}(a,b)=1$ and $c\vert a+b$ together imply that $\operatorname{gcd}(c,a)=\operatorname{gcd}(c,b)=1.$ The hint suggests you divide this task into two hopefully easier ones in the following way:
Suppose that $\operatorname{gcd}(a,b)=1,$ $c\vert a+b$ and that the numbers $\operatorname{gcd}(c,a)$ and $\operatorname{gcd}(c,b)$ are equal to each other, calling their value $d.$ Prove that $d$ must then be $1.$
Let $\operatorname{gcd}(a,b)=1$ and $c\vert a+b.$ Show that this implies that $\operatorname{gcd}(c,a)=\operatorname{gcd}(c,b).$
Upon proving these two individual statements, you are done because combining them together implies the statement you originally had to prove.
But it seems to me from the original question that this much is clear to you, although you cite it as explanation why you find this approach invalid. I hope things are clearer now, but if not, maybe try to elaborate a bit on just what it is you find concerining.