$[0,1]^{[0,1]}$ is separable This is from Dudley´s book: Let $I:=[0,1]$ with the usual topology. Let $I^I$the set of all the functions from $I$ to $I$ with the product topology. 
a) $I^I$ is separable. Hint: Consider function that are finite sums $\sum a_i1_{j(i)}$, where $a_i\in \mathbb{Q}$ and the $j(i)$ are intervals with the rational endpoints.
b) Show that $I^I$ has a subset which is not separable with the relative topology.
Sketch proof:
a)  Let $A$ be the set of all the functions as described in the hint. We have to show that $A$ is indeed countable. There are countable many intervals with  rational endpoints  since each such interval can be identified with the pair $\langle a,b\rangle\in \mathbb{Q}\times \mathbb{Q}$ (for $a,b$ the endpoints) and the former is countable. Then any finite sequence of such intervals is countable. Since for such sequence of intervals we can always have a countable number of functions it follows that the family $A$ must be countable. 
To conclude it will suffice to show that for any non-empty basic open set $U$ the intersection of $A$ and $U$ is non-empty. Let $U$ be a basic open set, then 
$$U=\bigcap_{\{i\in F: F \text{ finite }\}}\text{pr}_i^{-1}(U_i)=\{f\in I^I:f(x)\in U_x\text{ and } x\in F \}$$
We take disjoint intervals with rational endpoints s.t. $x\in j(x)$ and also since $U_x\subset I$ for denseness of $\mathbb{Q}$ we choose $q_x\in {U_x}\cap \mathbb{Q}$. Hence $\sum_{x\in F} q_x1_{j(x)}$ is in $A\cap U$. 
For b) we have to show that $\{0,1\}^I$ is not separable with respect to the relative topology... 
The above argument is OK? , b) could someone give me a hint?
Thanks in advance :)
 A: For the first part, I wrote this proof (written down differently) as this answer, so you could compare. It's essentially the same.
For (b), the subset $\{0,1\}^I$ is not going to work, as this is also separable; a generalisation of the argument from (a) will show that any product of at most $|[0,1]|$ (i.e. continuum many) separable spaces will be separable, so certainly $\{0,1\}^I$. 
But consider the set $C = \{\chi_p: p \in [0,1]\}$, where $\chi_p: [0,1] \rightarrow [0,1]$ is defined by $\chi_p(x) = 0$ for $x \neq p$, $\chi_p(p) = 1$. 
Then the open sets $U_p = (\pi_p)^{-1}[(\frac{1}{2},1]] = \{f \in [0,1]^{[0,1]}: f(p) \in (\frac{1}{2},1] \}$ is basic open in $[0,1]^{[0,1]}$ and is such that $U_p \cap C = \{\chi_p\}$, which shows that $C$ is discrete as a subspace, so cannot be separable, as it has size $|[0,1]|$. (Note also that adding the constantly $0$ function to $C$ gives us a copy of the one-point compactification of a discrete space of size $|[0,1]|$. This gives us a compact subspace that is non-separable.)
