Ergodic without atoms implies completely conservative?

The desired statement is entirely measure-theoretic, so it's not really necessary to think about the topology of $X$. By standard descriptive set theory, there exists a Borel linear ordering $\preceq$ on $X$. (This means that the graph of $\preceq$ is a Borel subset of $X \times X$.)

Suppose toward a contradiction that $\mu$ is nonatomic and the $\Gamma$-action is ergodic but that there exists a wandering set $W$ of positive measure. Without loss of generality we may assume that the intersection of every $\Gamma$-orbit with $W$ is finite. The ordering $\preceq$ restricts to a linear ordering on each $\Gamma$-orbit. If a $\Gamma$-orbit $\mathcal{O}$ intersects $W$ then $\mathcal{O} \cap W$ is nonempty and finite, so that the $\preceq$-least element of $\mathcal{O} \cap W$ is well-defined. Since $\preceq$ is Borel, the set $V$ consisting of the $\preceq$-least element of each such set $\mathcal{O} \cap W$ is Borel. The set $V$ contains at most one point from each $\Gamma$-orbit. Since $W \subseteq \bigcup_{\gamma \in \Gamma} V$ and $\mu$ is $\Gamma$-quasi-invariant it follows that $\mu(V) > 0$.

Since $\mu$ is nonatomic, there exists a Borel set $V_1 \subseteq V$ such that if we write $V_2 = V \setminus V_1$ then $\min(\mu(V_1),\mu(V_2)) > 0$. Let $\widetilde{V}_j = \bigcup_{\gamma \in \Gamma} \gamma V_j$. For each $j \in \{1,2\}$ the set $\widetilde{V}_j$ is $\Gamma$-invariant and since $\widetilde{V}_j$ contains $V_j$ we have $\mu(\widetilde{V}_j) > 0$. By ergodicity it follows that $\mu(X \setminus \widetilde{V}_j) = 0$ for each $j \in \{1,2\}$.

We claim that $\widetilde{V}_1$ and $\widetilde{V}_2$ are pairwise disjoint, which will lead to the contradiction $$0 < \mu(\widetilde{V}_2) \leq \mu(X \setminus \widetilde{V}_1) =0$$

We will establish the claim by showing that that existence of $x \in \widetilde{V}_1 \cap \widetilde{V}_2$ leads to a further contradiction. Suppose that there exists such a point $x$. Then there exist $\gamma_1,\gamma_2 \in \Gamma$ with $\gamma_1x \in V_1$ and $\gamma_2x \in V_2$. Since $V_1 \cap V_2 = \emptyset$ we have $\gamma_1x \neq \gamma_2x$. Since $\gamma_1x = \gamma_1\gamma_2^{-1}\gamma_2x$ we see that $\gamma_1x$ and $\gamma_2x$ are in the same $\Gamma$-orbit. This cannot be the case since $\gamma_1x$ and $\gamma_2x$ are both elements of $V$ and $V$ contains at most one point from each $\Gamma$-orbit.

$\mathbf{Response\,\, to \,\,comment}$

The way to see that the set $V$ is Borel is to regard Borel sets as being defined by logical formulas. We say that a formula 'is Borel' if it defines a Borel subset of $X$. Countable existential quantifiers correspond to countable unions and not corresponds to complements. Thus these logical operations preserve the property of being a Borel formula.

It is implicit in the hypotheses that $\Gamma$ acts by Borel automorphisms. Since $\preceq$ is Borel, it follows that the predicate $\gamma x \prec x$ is Borel for any $\gamma \in \Gamma$. The set $V$ can be defined as $$\{x \in W: (\not \exists)(\gamma \in \Gamma)((\gamma x \in W) \wedge (\gamma x \prec x))\}$$.

First a general comment. Unfortunately, the categorical approach is completely missing in the way the measure theory is still taught nowadays, and the fact that there actually is only one "reasonable" purely non-atomic probability measure space (so called Lebesgue, Lebesgue--Rokhlin, or standard probability space) isomorphic to the unit interval endowed with the classical Lebesgue measure is not very widely known. Any Borel probability measure on a Polish space makes it a Lebesgue space, so that in particular all measure spaces one has to deal with in the Patterson theory are from this class.

The definition of a wandering set you are using is a non-standard one. According to the usual definition a measurable set is wandering if all its translates are pairwise disjoint mod 0 (i.e., the intersection with almost any orbit consists of at most one point). However, it is very easy to see that the existence of a wandering set $W$ in your sense implies the existence of a "usual" wandering set $W'$. Namely, realize your measure space as the unit interval, and then for each orbit take the minimal point from its intersection with $W$. This operation is clearly measurable and provides you with a measurable set $W'\subset W$, which is wandering in the usual sense.

You are asking why ergodicity implies complete conservativity. The reason (with the right definition of wandering sets) is very simple: the union of the translates of any non-trivial subset of a wandering set is obviously a non-trivial invariant set. Actually, more generally, the union of the free discrete ergodic components of the action is precisely the dissipative part of the Hopf decomposition. See Kaimanovich for more details and Grigorchuk - Kaimanovich - Nagnibeda for various examples concerning the relationship between dissipativity and minimality of boundary actions.