Let $R$ be a PID and $R'$ a Integral Domain and $R\subseteq R'$. Let $a,b,d \in R $ and $d$ is a gcd of $a$ and $b$ in $R$. Then $d$ is also a gcd of of $a$ and $b$ in $R'$.
Proof:
$d$ is a gcd of $a$ and $b$ in $R$ $\Leftrightarrow$ $(a,b)=(d)$. It is clear that $d$ is a divisor of of $a$ and $b$ in $R'$, we have to show that it is the Greatest divisor. Let $e\in R'$ be a divisor of $a$ and $b$ in $R'$. This means $\exists \, v,w \in R' \, : a=ve \, , b=we$. Because $(a,b)=(d) $ we can assume $\exists \, s,t\in R' \, : d=sa+tb = sev + tew = esv + etw = e(sv+tw)$ and $sv+tw\in R'$ since $s,t,v,w\in R'$. Let $c:=sv+tw$, then we found a $c\in R'$ with $ce=d$ so $e$ is a divisor of $d$ in $R'$.
Is this correct?
I've made a few edits to correct typos and confusing notations (be careful: if you use the old-fashioned notation $\left(x\right)$ for the ideal generated by a single element $x$, then you should not put extra parentheses around elements when you mean the elements themselves!). Now your proof is correct.
Note that you never use that $R'$ is an integral domain. It is a redundant assumption.