Let’s begin by completing your idea for the box topology. Let $U=\left[0,\frac23\right)^\omega$; this is open in the box topology. For $n\in\omega$ let
$$V_n=\left\{\langle x_k:k\in\omega\rangle\in I^\omega:x\in\left(\frac13,1\right]\right\}\;;$$
each $V_n$ is open in the box topology — indeed, even in the ordinary product topology — and $\mathscr{V}=\{U\}\cup\{V_n:n\in\omega\}$ is an open cover of $I^\omega$. Suppose that $\mathscr{V}_0\subseteq\mathscr{V}$ is finite. Let $F=\{n\in\omega:V_n\in\mathscr{V}_0\}$. Define a point $x=\langle x_n:n\in\omega\rangle\in I^\omega$ as follows:
$$x_n=\begin{cases}
0,&\text{if }n\in F\\
1,&\text{otherwise}\;.
\end{cases}$$
Then $$x\notin U\cup\bigcup_{n\in F}V_n\supseteq\bigcup\mathscr{V}_0\;,$$
so $\mathscr{V}_0$ doesn’t cover $I^\omega$, and $\mathscr{V}$ has no finite subcover.
For the uniform topology I think that another approach is easier. For $n\in\omega$ define the point $x^{(n)}=\langle x_k^{(n)}:k\in\omega\rangle\in I^\omega$ by
$$x_k^{(n)}=\begin{cases}
1,&\text{if }k=n\\
0,&\text{otherwise}\;.
\end{cases}$$
It’s not hard to show that $A=\{x^{(n)}:n\in\omega\}$ is an infinite subset of $I^\omega$ with no limit point in the uniform topology, which is enough to show that $I^\omega$ is not compact in the uniform topology. If you want to use open covers, for each $n\in\omega$ let $B_n$ be the open $\frac12$-ball centred at $x^{(n)}$. (In other words, $B_n=\prod_{n\in\omega}I_k$, where $I_k=\left[0,\frac12\right)$ if $k\ne n$, and $I_n=\left(\frac12,1\right]$.) Then show that $A$ is closed, and let $U=I^\omega\setminus A$; $\mathscr{U}=\{U\}\cup\{B_n:n\in\omega\}$ is then an open cover of $I^\omega$ with no finite subcover, since for each $n\in\omega$ the only member of $\mathscr{U}$ containing $x^{(n)}$ is $B_n$.