Let $T=\{t_1, t_2, t_3\}$. Construct $U=S \cup T$, then $|U| = k$. Pick an $x \in V -U$. By Fan Lemma, there is an x,U fan of size k, which gives 3 internally disjoint path $P_1: x \to t_1,P_2: x \to t_2, P_3: x\to t_3$.
Now let $G_1 = G-S$, which is a 3-connected graph by definition. By Menger’s thm, $\forall a,b \in T, \lambda_{G_1}(a,b) \ge 3$. Then there’s a path $Q_1: t_2 \to t_3$ that is internally disjoint from $P_3P_2$.
Similarly, there’s a path $Q_3: t_1 \to t_2$ that is internally disjoint from the path $P:=P_1P_3Q_1$. Then $P_1 P_3 Q_1 Q_3$ is a closed path (cycle) containing T.
cycle: C1 s-b-a-c-t-s with value 1
t-s path: P1 t-b-c-s with value 1, P2 t-s with value 1
s-t paths: P3 s-a-c-t with value 2, P4 s-a-t with value 0.5
Removal of C1 results in the middle figure, and further removing P1 gives the last figure.
Construct network G’ by adding source and sink vertex s and t. All edge capacities are 1.
The maximum flow of this network is $f=(s \to u_1 \to w_1 \to t) +(s \to u_2 \to w_2 \to t) +(s \to u_3 \to w_3 \to t)$ with value $M(f)=3$. It is maximum because in the residual network shown below, there is no augmenting path from s to t.
The maximum flow corresponds to the matching: $u_1 - w_1, u_2 - w_2, u_3 - w_3$. This matching is maximum because any larger matching, if exists, will corresponds to a feasible flow in G’ with larger value, yet f is already having the largest value in G’.
By Kőnig & Egerváry, $\beta(G) = \alpha’(G)$. Then a vertex cover with size 3 will be minimum. $\{w_1, w_2, u_3\}$ is a valid choice.
In 2-connected plane graph, every face is bounded by a cycle.