Petrank's proof for the APX-hardness of MAX k-VERTEX COVER on subcubic graphs
Here is the proof of Petrank's reduction. This was suggested to me by Édouard Bonnet. I still do not know whether this can be turned into an L-reduction.


  Suppose there is a PTAS algorithm $\mathscr{A}$ for MAX k VERTEX COVER, thus giving a $(1-\varepsilon)$-factor-approximate solution on a cubic graph $G=(V,E)$ with $|V|=n, |E|=m=3n/2$.
  We can use $\mathscr{A}$ to solve MIN VERTEX COVER on $G$ by invoking it for $k=1,2,\dots, n$, and keep the best of all the produced solutions. Assume that it is obtained for $k=k_0$.
  We claim that $\mathscr{A}$ for $k=k_0$ produces a solution where $k_0$ vertices cover at least $(1-\varepsilon)m$ edges. Indeed, when $k=opt_{VC}(G)$ (where $opt_{VC}(G)$ denotes the minimum solution size for MIN VERTEX COVER on $G$), at least $(1-\varepsilon)m$ edges must be covered by the result of $\mathscr{A}$. Since the result of $\mathscr{A}$ for $k=k_0$ is at least as good as this, the claim follows.
  
  Thus, at most $\varepsilon \cdot m$ edges are not covered. We construct a set $C$ with the $k_0$ vertices selected by $\mathscr{A}$, to which we add one arbitrary vertex of each of the remaining $\varepsilon \cdot m$ uncovered edges.
  This set $C$ is a vertex cover for $G$, and its size is $k_0 + \varepsilon \cdot m$. 
  However, in a cubic graph, $m \leq 3k_0$ (in any optimal solution for MIN VERTEX COVER, each vertex can cover at most 3 edges). 
  Therefore, $|C| \leq k_0 + \varepsilon \cdot 3k_0 = k_0(1+\varepsilon')$, by setting $\varepsilon' = 3\varepsilon$. This contradicts the APX-hardness of MIN VERTEX COVER on cubic graphs.