It is well known that the space $\{0,1\}^{\kappa}$ satisfies the countable chain condition. Recall that a topological space $X$ satisfies the countable chain condition if and only if every collection $\mathcal{A}$ of pairwise disjoint open sets is countable. However, it is easy to show that the surjective continuous image of a space satisfying the countable chain condition must also satisfy the countable chain condition, so the only possible spaces which are images of some $\{0,1\}^{\kappa}$ satisfy the countable chain condition. However, there are plenty of compact Hausdorff spaces that do not satisfy the countable chain condition such as $\beta\mathbb{N}\setminus\mathbb{N}$ or $[0,1]\times[0,1]$ with the order topology inherited from the lexicographic ordering.

To see that $\{0,1\}^{\kappa}$ satisfies the countable chain condition, endow $\{0,1\}^{\kappa}$ with the infinite product measure $m$ where each $\{0,1\}$ is given the measure $\mu$ such that $\mu(\{0\})=\frac{1}{2}$ and $\mu(\{1\})=\frac{1}{2}$. Then $m$ extends to a Borel measure $\overline{m}$ on $\{0,1\}^{\kappa}$ where $\overline{m}(U)>0$ for each non-empty open set $U$. However, there cannot be an uncountable collection $\mathcal{A}$ of disjoint open subsets of $X$ since the union of any uncountable pairwise disjoint collection of open subsets of $X$ would have infinite measure.

General topology. $\endgroup$ – Tomasz Kania Mar 18 '15 at 16:51