HINT: Essentially the same hint that I gave for this question works here. For each positive integer $n$ let $\mathscr{U}_n=\left\{B\left(x,\frac1n\right):x\in X\right\}$; this is an open cover of $X$, so it has a countable subcover $\mathscr{B}_n$. Consider $\mathscr{B}=\bigcup_{n\in\Bbb Z^+}\mathscr{B}_n$.
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.
Best Answer
HINT: For each positive integer $n$ let $\mathscr{U}_n=\left\{B\left(x,\frac1n\right):x\in X\right\}$; this is an open cover of $X$, so it has a finite subcover $\mathscr{B}_n$. Consider $\mathscr{B}=\bigcup_{n\in\Bbb Z^+}\mathscr{B}_n$.