Compact topological space not having Countable Basis?

A direct argument showing that $[0,1]^I$ is not second-countable for uncountable $I$ is not too difficult to drum up.

First consider the standard basis $\mathcal{B}$ for the product topology: all products of the form $\prod_{i \in I} U_i$ where each $U_i$ is open in $[0,1]$ and $U_i = [0,1]$ for all but finitely many $i \in I$.

A nice result is that given any basis of a topological space, you can always find a subset of that basis which itself is a basis, and of minimum possible size. So, if $[0,1]^I$ were second-countable, there would be a countable subset of $\mathcal{B}$ which is also a basis.

Letting $\mathcal{B}^\prime = \{ U^{(k)} : k \in \mathbb{N} \}$ be any countable subset of $\mathcal{B}$, for each $k \in \mathbb{N}$ we can write $U^{(k)} = \prod_{i \in i} U^{(k)}_i$ as above. For each $k$ let $I_k = \{ i \in I : U^{(k)}_i \neq [0,1] \}$. Each $I_k$ is finite, and so $\bigcup_{k \in \mathbb{N}} I_k$ is countable. Thus there is some $i_* \in I \setminus \bigcup_{k \in \mathbb{N}} I_k$. Letting $V_{i_*} = [0, \frac{1}{2} )$, and $V_i = [0,1]$ for all $i \neq i_*$, consider the open set $V = \prod_{i \in I} V_i$. It is not difficult to show that $U^{(k)} \not\subseteq V$ for all $k \in \mathbb{N}$, and so $\mathcal{B}^\prime$ cannot be a basis!

(By making the appropriate changes to this proof, you can show that the product of uncountably many spaces that don't have the trivial topology is not second-countable.)

For another basic example of a compact space which is not second-countable, consider the ordinal space $\omega_1+1 = [0, \omega_1]$ (where $\omega_1$ is the least uncountable ordinal). With a little knowledge of ordinals it is not too difficult to show that this space is not first-countable (there is no countable base at $\omega_1$), and therefore is not second countable.

Here's a very easy way to check that certain spaces cannot have a countable basis (though it will not detect all spaces that lack a countable basis). Note that if $\mathcal{B}$ is a basis for a topological space $X$, then for every open set $U\subseteq X$ there exists a subset $\mathcal{A}\subseteq\mathcal{B}$ such that $U$ is the union of the elements of $\mathcal{A}$. This means that there can be at most $2^{|\mathcal{B}|}$ different open subsets of $X$. So if $X$ has a countable basis, it can have at most $2^{\aleph_0}$ open subsets.

This makes it easy to find compact spaces with no countable basis. For instance, let $X$ be any compact $T_1$ space with more than $2^{\aleph_0}$ points (e.g., a product of copies of $[0,1]$, as long as you have enough copies that the cardinality of the product is greater than $2^{\aleph_0}$). Then $X$ cannot have a countable basis, since for each $x\in X$ there is a distinct open set $X\setminus\{x\}$.