Q:

Every vertex cover is a dominating set

Suppose $G$ is a connected graph and $S$ is a vertex cover. Prove that $S$ is also a dominating set.

Can I get some help with proving this? I know that a dominating set in an undirected graph $G=(V,E)$ is a subset $S$ subset of $V$ such that every vertex is either in $S$ or has a neighbor in $S$. 
I know that a vertex cover is always going to be a dominating set because a vertex cover set is such that every edge has a vertex in the set, which means that a vertex not in the set (so on the other end of the edge) must have a neighbor in the set in order for all edges to be covered...
I just don't know how to write this as a proof.

A:

Here is a correct version of your result:

If a graph contains no isolated vertices, then every vertex cover is a dominating set.

For the proof, let $G$ be a graph with no isolated vertices, and let $S$ be a vertex cover. In order to show that $S$ is a dominating set, we need to show that every vertex not in $S$ has a neighbor in $S$. Suppose therefore that $v \notin S$. Since $G$ has no isolated vertices, $v$ has some neighbor $u$. Since $S$ is a vertex cover, it must contain one of $v,u$. Since $v \notin S$, necessarily $u \in S$. Therefore $v$ has a neighbor $u \in S$.
A stronger version of this result states that the condition of having no isolated vertices is tight.

Let us call a graph good if every vertex cover is a dominating set.
A graph is good iff it contains no isolated vertices.

We already saw one direction. Now suppose $G$ is a graph containing some isolated vertex $v$. Take an arbitrary vertex cover $S$. Since $v$ is isolated, $S \setminus \{v\}$ is also a vertex cover, but it is not a dominating set, since $S \setminus \{v\}$ doesn't contain $v$ or any of its neighbors (since it has none!).

