Let G be a multipartite graph on r classes, each containing k vertices, such that there is no independent set which contains at least one vertex from each class. I believe such graphs contain a complete bipartite graph $K_{f(k),f(k)}$ for each fixed r, with f an increasing (possibly even linear) function, but the proof eludes me so far. Google search led me to extremely results about Ramseytype results for bipartite subgraphs, but not complete ones. Any insights are much appreciated.
Large bicliques in rpartite graphs containing no independent sets having one vertex from each class

I believe I can prove this with a standard Ramseytype argument, though f will grow slower than linear. You'll need the following useful lemma.
Proof of lemma goes by induction. Choose an upper vertex u. If this vertex is linked to at least $ R_1(n_0  1, n_1, m_0, m_1) $ lower vertices and there are enough upper vertices, then there's either a large empty induced subgraph, or a complete bipartite subgraph that's large enough if you include u. Similarly, if there are at least $ R_1(n_0, n_1, m_0  1, m_1) $ lower vertices u is not linked to, you've won. Thus, $$ R_1(n_0, n_1, m_0, m_1) := R_1(n_0  1, n_1, m_0, m_1), + R_1(n_0, n_1, m_0  1, m_1), $$ $$ R_0(n_0, n_1, m_0, m_1) := 1 + \max(R_0(n_0  1, n_1, m_0, m_1), R_0(n_0, n_1, m_0  1, m_1)) $$ vertices are enough. The base case when $ n_0 = 0 $ or $ m_0 = 0 $ is trivial. Now for the theorem you are asking for, we can prove like this.
The case where $ P $ is the complete graph gives the theorem you're asking for:
Proof of lemma 2 goes by fixing $ r, f $, then taking induction on $ P $. The base case when $ P $ is an empty graph is trivial: $ m $ vertices in each class of the bipartite graph is enough. Otherwise, suppose we already have the induction statement for pattern $ P' $, which is $ P $ with the edge $ {i, j} $ deleted, and we found that $ k' $ vertices per class is enough. We have a large enough graph $ G $ (with at least $ k $ vertices in each class, $ k $ is to be determined later) that does not contain a $ K_{f,f} $ complete bipartite graph as a subgraph. We want to prove that $ G $ contains $ r $ vertices unlinked according to the pattern $ P $. Choose $ k = \max(R_0(f, f, k', k'), R_1(f, f, k', k')) $. Consider the induced subgraph of $ G $ made of only its classes $ i $ and $ j $. Both classes of this bipartite subgraph still has at least $ k $ vertices, and the subgraph still doesn't contain a $ K_{f,f} $ complete bipartite graph. Thus, you can apply lemma 1 with $ n_0 = n_1 = f $ and $ m_0 = m_1 = k' $ to find that this subgraph has an empty induced subgraph formed of $ k' $ vertices from each of the two classes. Now let $ G' $ be the induced subgraph of $ G $ that contains the $ k' $ vertices from class $ i $ and $ k' $ vertices from class $ j $ as chosen above, and all vertices from all other classes. Thus, $ G' $ has no edges between class $ i $ and $ j $. By the induction statement, we can choose vertices $ v_1, \dots, v_r $ from each class of $ G' $ such that they are unlinked accoring to pattern $ P' $, but as $ v_i $ and $ v_j $ are also unlinked, these vertices are also unlinked according to pattern $ P $. QED. 

