In a three-connected plane graph, face boundaries are exactly the induced cycles whose deletion leaves the graph connected

A facial boundary $C$ is a cycle by [L1]. It has no chord $xy$: the chord lies on the nonfacial side, and the chord together with each of the two $x$-$y$ arcs of $C$ is a polygon by [L3]. These two polygons bound disjoint subregions of the nonfacial side. Since the facial side contains no graph edge, a path from an internal vertex of one $x$-$y$ arc to an internal vertex of the other, avoiding $x,y$, would have to cross the chord. Thus $\{x,y\}$ would be a vertex cut, contrary to three-connectivity and [L2]. Hence $C$ is induced.

Conversely, let $C$ be induced and suppose $G-V(C)$ is connected or empty. If $C$ is not facial, both complementary regions from [L3] contain graph material. Inducedness rules out a chord as that material, so each side contains a vertex outside $C$. When $G-V(C)$ is connected, a path in it joins vertices on the two sides and must cross $C$, contradicting the plane embedding. If $G-V(C)$ is empty, the drawing consists only of $C$, and both sides are faces. Thus $C$ is facial.

Suppose $G-V(C)$ were disconnected, and let $D_1$ be one of its components. Every component has at least three neighbours on $C$, since one or two such neighbours would be a vertex cut contradicting [L2]. Because $C$ bounds a face, every component lies in the closed complementary side of $C$, which [L3] presents as a region bounded by $C$. Let $a_1,\ldots,a_k$ with $k\ge3$ be the attachments of $D_1$ in cyclic order on $C$, and let $S$ be a spanning tree of $D_1$ together with one edge to each $a_i$. Then $S$ is connected, lies in that side, and meets $C$ exactly in $\{a_1,\ldots,a_k\}$, so by [L3] the set $S\cup C$ divides the side into exactly $k$ regions, the $i$th bounded by the arc of $C$ from $a_i$ to $a_{i+1}$ carrying no further attachment of $D_1$ together with two paths of $S$. Every other component is connected and meets neither $S$ nor $C$, so each lies inside a single one of those open regions; choose one, say the $i$th, that contains a component, and put $x:=a_i$ and $y:=a_{i+1}$. This is where the alternation of attachments is not enough: two components may share all three of their attachments, in which case no four attachments alternate, and the region argument rather than the crossing argument is what confines them. Let $U$ consist of the open arc of $C$ strictly between $x$ and $y$ together with every component lying in the $i$th region. Distinct components of $G-V(C)$ are non-adjacent, and each component in that region has all of its neighbours on the closed arc from $x$ to $y$, while $D_1$ has no attachment strictly inside that arc. A vertex of the open arc has no neighbour elsewhere on $C$, because step 1.1 has already shown $C$ induced, so $C$ has no chord; and any component adjacent to such a vertex has an attachment strictly inside the arc, hence lies in the $i$th region and is already in $U$. Hence the only neighbours of $U$ outside $U$ are $x$ and $y$, so $\{x,y\}$ is a vertex cut separating $U$ from $D_1$, contradicting [L2]. Thus $G-V(C)$ is connected when nonempty.

Steps 1.1 and 2.1 prove the forward implication, and step 1.2 proves the reverse, including the empty-deletion boundary.