Does any subset of $\beta\omega$ of cardinality $\mathfrak{c}$ have a weak P-point in its closure? I believe that anyone who can answer this question knows the terminology, but $\beta\omega$ is the Čech-Stone compactification of the integers and a point $p$ is a weak P-point in a space $X$ iff it is not in the closure of any countable subset of $X-\{p\}$. I admit to lacking intuition with $\beta\omega$ and maybe this question has a trivial answer, apologies if it is the case.
My motivation is the following.
I want to find a countably compact space $X$ and a surjective map $f:X\to[0,1]$ such that $f(C)\not=[0,1]$ if $C\subset X$ is compact. There might be a simple example that escaped me, but if the question in the title has a positive answer, taking $X$ to be $\beta\omega$ minus its weak P-points yields such a space: 
-It is countably compact because $\beta\omega$ is compact, hence any countable subset has a cluster point which cannot be a weak P-point,
-It maps onto $[0,1]$ because we may map $\omega$ onto the rational numbers in $[0,1]$ and extend the function to all of $\beta\omega$. The image of $X$ is countably compact, hence compact in $[0,1]$ and is thus all of $[0,1]$.
-If $f(E)=[0,1]$ with $E\subset X$, $E$ has cardinality $\mathfrak{c}$ and its closure non-compact.
But maybe I am trying to kill a mosquito with an atomic bomb and a much simpler example exists.
 A: The answer is no, because every infinite closed subset of $\beta \omega$ has cardinality $2^\mathfrak{c}$. So for example, if $\{x_1,x_2,\dots\}$ is any countably infinite set of non-weak-$P$-points, then its closure $X$ contains no weak $P$-points either. Any $Y \subseteq X$ of cardinality $\mathfrak{c}$ provides a negative answer to the question.
A: To get a countably compact space $X$ such that there is a continuous surjection $f \colon X \to [0,1]$ but if $C$ is a compact subset of $X$, $f(C) \ne [0,1]$, let $X$ be a Bernstein-type set in $\omega^*$, that is, a subset of $\omega^*$ such that neither $X$ nor $\omega^* \setminus X$ contains an infinite compact set. (This can be done by mimicking the construction of a Bernstein set in $\mathbb R$: Well-order the infinite compact subsets of $\omega^*$ and inductively choose pairs of points one of which is in $X$ and the other of which is not in $X$.) Then $X$ is countably compact because any infinite subset has a limit point in $X$--all limit points cannot be in $\omega^* \setminus X$. It follows from the fact that fibers of continuous functions from $\omega^*$ to $[0,1]$ have non-empty interiors that if $F$ maps $\omega^*$ onto $[0,1]$, then the restriction $f$ of $F$ to $X$ maps $X$ onto $[0,1]$. Finally, since no compact subset of $X$ is infinite, no compact subset of $X$ maps onto $[0,1]$ by any function.
