Proof that the ideal of set of polynomials is generated by its gcd Theorem: Given $\{f_i\}_{1 \leq i \leq n}$, $f_i \in \mathbb{K}[x]$.  Then the monic generator $f$ of the ideal $\langle \{f_i\} \rangle$ is $f = \gcd \{ f_i \}$.
In other words: $\langle \{f_i\} \rangle = \langle \gcd \{f_i\} \rangle$
My try at a proof: Let $d = \gcd \{f_i\}$.    
$\subseteq$) If $g \in \langle \{f_i\} \rangle$, $g = \sum_i q_i f_i$ for some $q_i \in \mathbb{K}[x]$.  As $d | f_i \ \forall i$, $f_i = r_i d$ then $g = \sum_i q_i r_i d$, and that means that $g = d \sum_i q_i r_i$ that is $d | g$, $g \in \langle d \rangle$
$\supseteq$) Let $g \in \langle d\rangle$.  We know that $d = \sum_i q_i f_i$.  And that $g = d q$.  Then $dq = g = \sum_i q_i q f_i$.  Then $g \in \langle \{f_i\} \rangle$.
I would like to know if my proof is correct.  What puzzles me most is that I hadn't used that d is the greatest common divisor, just that it divides the other polynomials.
 A: My impression is that the point is showing that $\langle d\rangle\subseteq\langle f_1,f_2,\dots,f_n\rangle$ which is equivalent to
$$
d=\sum_{k=1}^n q_kf_k.
$$
The impression follows from the fact that once you know this the entire proof is trivial. Let's take another path.
You should know that $\langle f_1,f_2,\dots,f_n\rangle=\langle g\rangle$ for a unique monic polynomial $g$. Let's show that $g$ is the (monic) greatest common divisor of $f_1,\dots,f_n$. (By general theory about $\mathbb{K}[x]$, $g$ is the monic polynomial of least degree in $\langle f_1,f_2,\dots,f_n\rangle$.)
First of all, from $f_k\in\langle g\rangle$ we deduce that $g$ divides $f_k$ $(k=1,2,\dots,n)$. Now, let $h$ be a polynomial that divides $f_k$ $(k=1,2,\dots,n)$ and let us show that $h$ divides $g$.
Since $g\in\langle f_1,f_2,\dots,f_n\rangle$, we can write $g=\sum_{k=1}^n q_kf_k$; also $f_k=p_kh$, so
$$
g=\sum_{k=1}^n q_kp_kh=\biggl(\sum_{k=1}^n q_kp_k\biggr)h
$$
and $h$ divides $g$ as requested.
Therefore $g$ is the greatest common divisor of the given polynomials.
A: The proof is by and large correct; if I would write it myself I would present some things differently, but I would grade it as correct. 
To adress your specific doubt. 
The first part of your argument is a special case of the fact that the ideal generated by the $f_i$ is contained in the ideal generated by $h$ for $h$ being some divisor common to the $f_i$; thus here you really only need that it is a divisor common to the $f_i$. 
The second part of your argument uses the facts that a greatest common divisor is an element of the ideal generated by the $f_i$, and only this property is important for this part (that this element is a common divisor is not used). 
As mentioned in a comment, it is not true for each common divisor that it is an element of the ideal generated by the $f_i$; for this you need it is the  greatest.
