You can find an example of a cascade which has topological entropy equal to zero, and which has a finite closed cover having entropy equal to $\log 2$ in the article "The Product Theorem for Topological Entropy" by L.W. Goodwyn.
Let $\hslash (T) = \sup h(\alpha, T)$, where the supremum is taken over all finite closed covers. We have that $\hslash (T)$ may not be equal to $h(T)$. In fact, that's what usually happens.
I don't know if the example presented in the dissertation is the same as in the article. But the idea of the example should be the same.
For a positive integer $m$, let $C_m$ denote the set of all bisequences $(\ldots, x(- 1), x(0), x(1),\ldots)$ of points in $I^m$, i.e., $C_m = (I^m)^{\mathbb{Z}}$. We let $C_m$ have the product topology and we define $\sigma_m: C_m \to C_m$ by
$$\sigma(x)(n)=x(n+1),$$
and let $\pi_m :C_m \to I^m$ be the projection $\pi_m(x) = x(0)$ for $x \in C_m$.
The example you are looking for will be a subcascade of the sequence cascade $(C_1, \sigma_1)$. For each positive integer $m$, we define $X_m$ to be the set of all bisequences $x= (\ldots, -x(-1), x(0), x(1),\ldots) \in C_1$ satisfying the requirement that there are no more than $m$ integers $i$ such that $|x(i) -\frac{1}{2}|>1/m$. Every $X_m$ is closed and $\sigma_1$-invariant. Define
$$X= \bigcap_{m=1}^{\infty} X_m,\qquad T=\sigma_1|_{X}\quad\text{and}\quad \pi=\pi_1|_X.$$
So you have $h(\alpha,T)=0$ (check), the topological entropy of $T$ is zero. To see that $\hslash (T) > h(T)$, define $F_0 = \{x \in X: x(0) \in [0,\frac{1}{2}]\}$ and $F_1 = \{x\in X : x(0) \in [\frac{1}{2},1]\}$, and let $\beta=\{F_0,F_1\}$. We have $h(\beta,T)=\log 2$.