This should be a simple question, but I am a little bit confused.

A proof on page 556 of Algorithm Design says:

“Let $ e=(u, v)$ be any edge of $ G$ . The graph $ G$ has a vertex cover of size at most $ k$ if and only if at least one of the graphs $ G-\{u\}$ and $ G-\{v\}$ has a vertex cover of size at most $ k-1$ .”

But when I try the following example where the black ball shows a vertex cover, it seems to be faulty statement. Because $ k$ remains as $ k$ (not $ k-1$ ) even after deletion of vertex $ u$ . Moreover, if I delete the vertex $ v$ , there would be no vertex cover left.

What is wrong about my deduction and example?