$\newcommand{\diam}{\operatorname{diam}}$I have the following theorem:
Let $X$ be a compact metric space, $T:X \to X$ be continuous and $(\alpha_n)$ be a set of open covers such that $\diam(\alpha_n)\to 0.$ Then $$h^*(T)=\lim_{n \to \infty} h^*(T ,\alpha_n).$$
From this, I want to conclude:
$$h^*(T)=\lim_{\epsilon\to0} \{\sup h^*(T,\alpha):\diam(\alpha)<\epsilon\}$$
Here, $h^*(T,\alpha)$ is the entropy of $T$ with respect to $\alpha$ and $h^*(T)=\sup\{h^*(T,\alpha):\alpha \text{ is an open cover} \}.$
My attempt:
Let $\epsilon>0$ be given and let $(\alpha_n)$ be an open cover such that $\diam(\alpha_n)=\frac{1}{n}. $ Then there exists $n_0\in \mathbb{N}$ such that $$\diam(\alpha_n)=\frac{1}{n}<\epsilon$$ for all $n \geq n_0.$ Thus, $$\sup\{ h^*(T,\alpha):\diam(\alpha) <\epsilon\} \geq h^*(T,\alpha_n)$$ for all $n \geq n_0.$ It follows that $$h^*(T)\geq\lim_{\epsilon\to0} \{\sup h^*(T,\alpha):\diam(\alpha)<\epsilon\}\geq \lim_{n \to \infty}h^*(T,\alpha_n)=h^*(T).$$
Is the above argument correct?