Can we prove that all CFLs can be recognized by a Turing Machine in polynomial time? This question came up while a group of students at my school were studying for our qualifying exams.  The question on an old exam was,


  Consider the following six classes of languages: Context free (CFL), Regular(REG), P, NP, Recursive(R), and Recursively enumerable(RE). (1) indicate the relationships between these six classes e.g., by drawing a Venn diagram. (2) name the minimum type of machine needed to recognize languages in the different classes.


OK, so we know that REG$\subseteq$CFL, either P$\subseteq$NP or P=NP, NP=RE, and P=R.

We also know that the minimum machine type for each language class is REG:DFA, CFL:PDA, P/R:DTM, NP/RE:might be NTM.

We know that a Turing machine is more powerful than a pushdown automata, but we can't find anything that actually proves that all CFLs can be recognized in polynomial time, or think of a way to prove it ourselves.  When I asked the teacher who originally wrote the problem about this, he didn't have a ready answer.
There are several misconceptions in this question and too many to cover in a comment.


We do not know that $\mathrm{P}=\mathrm{R}$ or that $\mathrm{NP}=\mathrm{RE}$. In fact, we know that both of these things are false. The time hierarchy theoreom tells us that both $\mathrm{P}$ and $\mathrm{NP}$ are strictly contained within the recursive languages.
$\mathrm{P}$ is not the class of languages accepted by any old deterministic Turing machines: they're DTMs running in polynomial time. Similarly, $\mathrm{NP}$ languages are accepted by NTMs running in polynomial time.
Without time bounds, the class of deterministic TMs and the class of nondeterministic TMs accept exactly the same languages: you can simulate nondeterminism deterministically but exponentially slower.
Also, we know that $\mathrm{REG}\subsetneq\mathrm{CFL}$, which is consistent with what you wrote but a stronger statement. Exercise: can you give an example of a context-free language that is not regular?
Another nitpick: you write "$\mathrm{P}\subseteq\mathrm{NP}$ or $\mathrm{P}=\mathrm{NP}$"; the second statement is already included in the first.