Let $(\Omega,\mathcal A)$, $(E,\mathcal E)$ be measurable spaces, $I$ be a nonempty countable set, $(X_i)_{i\in I}$ be an $(E,\mathcal E)$-valued process on $(\Omega,\mathcal A)$ and $$\kappa_\omega(B):=\sum_{i\in I}\delta_{X_i(\omega)}(B)\;\;\;\text{for }(\omega,B)\in\Omega\times\mathcal E.$$ Assume $\{x\}\in\mathcal E$ for all $x\in E$. Fix $\omega\in\Omega$. Note that $$\kappa_\omega\{X_i(\omega)\}=\left|\left\{j\in I:X_i(\omega)=X_j(\omega)\right\}\right|\ge1>0\tag1$$ and hence $X_i(\omega)$ is an atom of $\kappa_\omega$ for all $i\in I$.
How can we show that $\kappa_\omega$ is purely atomic$^1$?
Can we maybe even show that $D_\omega=\{X_i(\omega):i\in I\}$?
$^1$ i.e. $$D_\omega:=\left\{x\in E:x\text{ is an atom of }\kappa_\omega\right\}$$ is countable and $$\kappa_\omega(D_\omega^c)=0.$$
You seem to be on the right path with your intuition. Indeed, all atoms for the measure $\kappa_\omega$ are on the form $X_i(\omega)$ for some $i\in I$. To show that $\kappa_\omega$ is purely atomic you must verify the implication: $$ B\in\mathcal{E},\kappa_\omega(B)>0\Rightarrow B\text{ contains an atom.} $$ You have already shown that $X_i(\omega)$ is an atom for any $i\in I$ so it seems reasonable to look for an $i_0\in I$ such that $X_{i_0}(\omega)\in B$. Try to write down $\kappa_\omega(B)$ and see if you can deduce the existence of such an index $i_0$.
With the above approach in mind you should be all set to also prove that $D_\omega=\{X_i(\omega):i\in I\}$.