Every vertex cover is a dominating set
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!).