Finding a set which dominates the Minimum Dominating Set Given an unweighted, undirected graph, a dominating set $S$ is a set of nodes such that every node is in $S$ or adjacent to a node in $S$.

The dominating set problem is NP-hard, but I am considering a new variation.
Given $k$, the size of the minimum dominating set, I need to find a set of nodes $T$ of size at most $k$ such that there exists a minimum dominating set $S$ in the graph which is dominated by $T$. That is, every any node in $S$ should be in $T$ or adjacent to a node in $T$.

My attempt: 
The following is an algorithm I came up with that I have not proven or found a counterexample for.
At the start, all nodes are unmarked.

1) Pick an unmarked node $v$ as a center and mark everything within 2 hops of $v$. EDIT: Instead of picking an arbitrary unmarked node in this step, select the node that would cause the least number of nodes to be marked.

2) Repeat step 1 until all nodes are marked. (Note, this will never place more than $k$ centers)

3) If we have picked less than $k$ centers, keep placing centers on undominated nodes until we have placed $k$ centers.

I have proved that the above algorithm finds a set of centers which dominates a minimum dominating set, under the condition that we don't place any centers in step 3. That is, the algorithm works if steps 1 and 2 manage to place exactly $k$ centers.

Any ideas for whether this algorithm works or a counterexample?
The algorithm previously proposed in the post (before the current edit), does not work.

Here are the problem and algorithm proposed in that version of the post:

input: graph $G=(V,E)$ and the minimum size $k$ of any dominating set in $G$

output: a set of $k$ nodes that dominates a dominating set of size $k$


  algorithm$(G,k)$:
  
  At the start, all nodes are unmarked.
  
  1) Pick an unmarked node $v$ as a center and mark everything within 2
  hops of $v$.  [This step was edited and differs in the current post.]
  
  2) Repeat step 1 until all nodes are marked. (Note, this will never
  place more than $k$ centers)
  
  3) If we have picked less than $k$ centers, keep placing centers on
  undominated nodes until we have placed $k$ centers.





  Lemma. The algorithm above doesn't work.


Proof. Consider the counter-example $G=(V,E)$ where 


$V=\{a, b, c, d, e, f\}$
$E=\{(a, b), (a, c), (a, d), (b, e), (c, f), (d, f)\}.$




The (unique) minimum dominating set is $\{b, f\}$, with size $k=2$.

The algorithm can place a center on $v=a$ in Step 1 (marking all nodes), then place a center on $e$ in Step 3.  This returns a set $\{a, e\}$ of two centers, neither of which dominates $f$ (which is in the only dominating set of size $2$). $~~~\Box$