Let $X = \{x_1, x_2, \cdots, x_n\}$ be a $G$-transitive set and $H = G_{x_1}$.
It is required to prove that there are $g_1, g_2, \cdots, g_n \in G$ with $g_i x_1 = x_i$ such that $g_1 H, g_2 H, \cdots, g_n H$ are all the left cosets of $H$ in $G$.
What has been done:
Let $H = G_{x_1}$. Since $X$ is $G-$transitive set, for $x_1, x_i \in X$, there is some $g_i \in G$ such that $g_i x_1 = x_i$, for every $i \in J:=\{1,2, \cdots, n\}$.
Let $g \in G$. Remember that $\text{Orb}(x_1) = \{gx_1: g \in G\} \subseteq X$, but $\text{Orb}(x_1) = X$, since $X$ is a $G-$transitive set. Thus, $gx_1 = x_j$ for a unique $j \in J$, which implies that $g \in g_j H$.
By the other hand, let $i, j \in J$ such that $i \neq j$. Thus, $g_i \neq g_j$. Let's suppose that $g_j \in g_i H$, which implies that $g_j = g_i h$, for some $h \in H \subseteq G$. Thus, $x_j = g_j x_1 = g_i h x_1 = x_i$, which is a contradiction, since $i \neq j$ and $x_i \neq x_j$. This means that $g_j \not \in g_i H$.
This means that $G$ can partitioned into $n$ disjoint and non-empty sets.
Therefore, $K = \{g_1H, g_2H, \cdots, g_n H\}$ is the set of all left cosets of $H$ in $G$, all distinct two by two.
Is this proof complete and correct?
For every $i=1,\dots,n$ you can surely find $g_i\in G$ such that $g_ix_1=x_i$.
Take $g\in G$ and the coset $gH$. We want to see that $gH=g_iH$, for some $i$.
The natural choice would be $g_i$, where $gx_1=x_i$. Since $gx_1=g_ix_1$ by definition, we see that $g^{-1}g_ix_1=x_1$, hence $g^{-1}g_i\in H$ and $g_i\in gH$. Therefore $g_iH=gH$.
As you see, the proof is much simpler than your attempt. And your final part is disputable: the fact you have partitioned $G$ into disjoint sets (assuming you actually proved it) doesn't imply that this partition is the same as with cosets.