$\{0,1\}^{\omega}$ is not compact $\{0,1\}^{\omega}$ is not compact. I proved a more difficult result that $[0,1]^{\omega}$ is not compact in the box topology and was told my proof was similar to this. However, I think it would be good practice to try this result as well. $\{0,1\}$ is in the discrete topology. Let $C$ be the collection of all points in $\{0,1\}^{\omega}$.  Consider the open cover of $\{0,1\}^{\omega}$ given by $\{\prod_{n \in \mathbb{N}}U_{nx}\}_{x \in C}$, where for each $x=(x_n)_{n \in \mathbb{N}}$ in $C$, a corresponding open set $\prod_{n \in \mathbb{N}}U_{nx}$ containing it is defined by
$$U_{nx}=\begin{cases}   \{1\}   &\text{ if } x_n=1 \\
          \{0\}  & \ \text{if} \ x_n=0\end{cases}$$
Then $x \in \prod_{n \in \mathbb{N}}U_{nx}$ and if $y \in C-\{x\}$ there is an index $k$ such that $x_k \notin U_{ky}$ and $x \notin \prod_{n \in \mathbb{N}}U_{ny}$. So if $\prod_{n \in \mathbb{N}}U_{nx}$ is taken out of the cover, the element $x \in \{0,1\}^{\omega}$ is not covered by $\{\prod_{n \in \mathbb{N}}U_{nx}\}_{x \in C}-\{\prod_{n \in \mathbb{N}}U_{nx}\}$. Since $\{\prod_{n \in \mathbb{N}}U_{nx}\}_{x \in C}$ is an infinite cover, and taking one element of this collection out results in a noncover, $\{0,1\}^{\omega}$ is not compact.
One thing I had difficulty showing is that the collection $\{\prod_{n \in \mathbb{N}}U_{nx}\}_{x \in C}$ actually covers $\{0,1\}^{\omega}$. How could I show that the union of elements in the cover is equal to $\{0,1\}^{\omega}$ explicitly? Also, would this be the correct way to do this proof? I modeled the solution after this solutions, in which I was provided help. Proving $[0,1]^{\omega}$ in box topology is not compact

That it covers is trivial: $x \in \prod_n U_{nx}=: U_x$ for each $x \in C$.
In fact $U_x=\{x\}$ is open in the box topology, showing that all points of $\{0,1\}^\omega$ are isolated. $C$ is infinite (uncountable) and discrete.
The cover $\{U_x\mid  x\in C\}$ is irreducible: we cannot omit any $U_y$ because then $y$ will not be covered anymore. So the cover has no proper subcovers at all.
Nitpick: you can just say $U_{nx}=\{x_n\}$ and be done.
