Yes, every separating graph has a spanning subgraph which is minimally separating. The proof uses the same idea as the Banakh–Petrov theorem.

Let $G=(V,E)$ be a separating graph. I will write $N(x)$ and $d(x)=|N(x)|$ for the neighborhood and the degree of a vertex $x$ in $G$, and I will write $N_1(x)$ and $d_1(x)$ for the neighborhood and the degree of $x$ in the spanning subgraph $G_1=(V,E_1)$ to be constructed in Step 1.

**Step 1.** Let $G_1=(V,E_1)$ be a maximal spanning subgraph of $G$ with maximum degree $\Delta(G_1)\le3$, and let $W=\{x\in V:d_1(x)=3\}$; thus every edge $e\in E\setminus E_1$ has at least one endpoint in $W$.

**Step 2.** We will now construct a set $E_2\subseteq E\setminus E_1$ such that $G_{1,2}=(V,E_1\cup E_2)$ is a separating graph, and $G_{1,2}-e$ is non-separating for each $e\in E_2$. In order to make $G_1$ a separating graph by adding new edges, we only have to worry about vertices $x$ such that either $d_1(x)\lt2$ or else $d_1(x)=2$ and $x$ is in a triangle which has at least two such vertices. We consider several cases. The locution "draw a new edge" shall mean "choose an edge $e\in E\setminus E_1$ and add it to $E_2$"; the set $E_2$ shall consist of all the new edges chosen in Step 2.

**Case I.** $d_1(x)=0$.

Draw two new edges joining $x$ to vertices in $W$.

**Case II.** $d_1(x)=d_1(y)=1$ and $xy\in E_1$.

Draw two new edges joining $x$ and $y$ to two distinct vertices in $W$.

**Case III.** $d_1(x)=1$ and there are vertices $y\in V\setminus W$ and $z\in W$ such that $xy,yz\in E_1$.

If possible, draw a new edge joining $x$ to a vertex in $W$ distinct from $z$. If that's not possible then draw two new edges, joining $x$ to $z$ and joining $y$ to another vertex in $W$.

**Case IV.** $d_1(x)=1$ and neither Case II nor Case III applies.

Draw a new edge joining $x$ to a vertex in $W$.

**Case V.** $d_1(x)=d_1(y)=2$ and there is a vertex $z\in W$ such that $xy,xz,yz\in E_1$.

Draw a new edge joining either $x$ or $y$ to another vertex in $W$.

**Case VI.** $d_1(x)=d_1(y)=d_1(z)=2$ and $xy,xz,yz\in E_1$.

Draw two new edges joining two distinct vertices in $\{x,y,z\}$ to vertices in $W$, not necessarily distinct.

Let $E_2$ be the subset of $E\setminus E_1$ consisting of all the new edges from Step 2. It is easy to see that the graph $G_{1,2}=(V,E_1\cup E_2)$ is separating, and for each $e\in E_2$ the graph $G_{1,2}-e$ is non-separating.

**Step 3.** We want to find a minimal set $F\subseteq E_1\cup E_2$ such that $(V,F)$ is a separating graph; equivalently, a maximal set $S\subseteq E_1\cup E_2$ such that $(V,(E_1\cup E_2)\setminus S)$ is a separating graph.

Call a set $S\subseteq E_1\cup E_2$ *good* if $(V,(E_1\cup E_2)\setminus S)$ is a separating graph, *bad* if $(V,(E_1\cup E_2)\setminus S)$ is not a separating graph. Plainly, a subset of a good set is good. We want to find a maximal good set.

**Claim.** Every bad set $S\subseteq E_1\cup E_2$ contains a finite bad set.

**Proof of Claim.** Suppose $S$ is a bad set. Since $\{e\}$ is bad whenever $e\in E_2$, we may assume that $S\subseteq E_1$. By the definition of a separating graph, there are vertices $x,y\in V$ such that $S$ contains a bad subset $S_0$ consisting of edges incident with $x$ or $y$, that is, $S_0\subseteq N_1(x)\cup N_1(y)$. But then $S_0$ is finite, since the graph $G_1$ is locally finite, being subcubic.

It follows from the Claim and Zorn's lemma that there is a maximal good set $S\subseteq E_1\cup E_2$, whence $(V,(E_1\cup E_2)\setminus S)$ is a spanning subgraph of $G$ which is minimally separating.

**Remark.** A minimally separating graph is triangle-free.

Suppose $G$ is a separating graph, and suppose $G$ contains a triangle with vertices $x,y,z$. At least two of the three vertices, say $x$ and $y$, have degree at least $3$. If $G-xy$ is not a separating graph, then there must be a vertex of degree $2$ which is adjacent to $x$ and $z$ or to $y$ and $z$; let's say $N(w)=\{x,z\}$. But now it's easy to see that $G-xz$ is a separating graph, so $G$ is not minimally separating.