[proofplan]
Let $N(\mathcal V)$ denote the least size of a finite subcover of a finite open cover $\mathcal V$. Define $a_n=\log N(\mathcal U_0^{n-1})$. A subcover for the first $n$ iterates and a pulled-back subcover for the next $m$ iterates combine to cover the first $n+m$ iterates, so $a_{n+m}\le a_n+a_m$. Fekete's lemma gives existence of $\lim a_n/n$ and identifies it with $\inf a_n/n$.
[/proofplan]
[step:Define the iterated covers and their covering numbers]
For $n\geq 1$, define the joined cover
\begin{align*}
\mathcal U_0^{n-1}:=\bigvee_{j=0}^{n-1}T^{-j}\mathcal U.
\end{align*}
Thus its elements are finite intersections
\begin{align*}
U_0\cap T^{-1}U_1\cap\cdots\cap T^{-(n-1)}U_{n-1}
\end{align*}
with $U_j\in\mathcal U$. Since $\mathcal U$ is finite, $\mathcal U_0^{n-1}$ is finite. Since $X$ is compact and $\mathcal U_0^{n-1}$ is an open cover of $X$, it has a finite subcover. Let
\begin{align*}
N(\mathcal U_0^{n-1})
\end{align*}
be the smallest cardinality of such a subcover. Define
\begin{align*}
a_n:=\log N(\mathcal U_0^{n-1}).
\end{align*}
Then $a_n$ is a finite non-negative real number for every $n\geq 1$.
[guided]
The cover $\mathcal U_0^{n-1}$ records which element of $\mathcal U$ contains $x$, which element contains $Tx$, and so on through $T^{n-1}x$. It is the join
\begin{align*}
\mathcal U_0^{n-1}:=\bigvee_{j=0}^{n-1}T^{-j}\mathcal U.
\end{align*}
Because $\mathcal U$ is finite, this joined cover is finite. Because $X$ is compact, every open cover has a finite subcover, so the minimal subcover size
\begin{align*}
N(\mathcal U_0^{n-1})
\end{align*}
is well-defined and finite. We set
\begin{align*}
a_n=\log N(\mathcal U_0^{n-1}).
\end{align*}
[/guided]
[/step]
[step:Prove subadditivity]
Fix $n,m\geq 1$. Let $\mathcal A\subseteq \mathcal U_0^{n-1}$ be a subcover of $X$ with
\begin{align*}
|\mathcal A|=N(\mathcal U_0^{n-1}),
\end{align*}
and let $\mathcal B\subseteq \mathcal U_0^{m-1}$ be a subcover of $X$ with
\begin{align*}
|\mathcal B|=N(\mathcal U_0^{m-1}).
\end{align*}
Consider the family
\begin{align*}
\mathcal C:=\{A\cap T^{-n}B:A\in\mathcal A,\ B\in\mathcal B\}.
\end{align*}
Every member of $\mathcal C$ belongs to $\mathcal U_0^{n+m-1}$, because $A$ controls the first $n$ iterates and $T^{-n}B$ controls the following $m$ iterates. Also $\mathcal C$ covers $X$: if $x\in X$, choose $A\in\mathcal A$ with $x\in A$ and choose $B\in\mathcal B$ with $T^nx\in B$; then $x\in A\cap T^{-n}B$. Hence
\begin{align*}
N(\mathcal U_0^{n+m-1})\leq |\mathcal C|\leq |\mathcal A|\,|\mathcal B|.
\end{align*}
Therefore
\begin{align*}
N(\mathcal U_0^{n+m-1})\leq N(\mathcal U_0^{n-1})N(\mathcal U_0^{m-1}).
\end{align*}
Taking logarithms gives
\begin{align*}
a_{n+m}\leq a_n+a_m.
\end{align*}
[guided]
A minimal subcover for the first $n$ iterates and a minimal subcover for the next $m$ iterates can be multiplied to cover the first $n+m$ iterates. Choose subcovers $\mathcal A$ and $\mathcal B$ with sizes
\begin{align*}
|\mathcal A|=N(\mathcal U_0^{n-1})
\end{align*}
and
\begin{align*}
|\mathcal B|=N(\mathcal U_0^{m-1}).
\end{align*}
For each $A\in\mathcal A$ and $B\in\mathcal B$, the set $A\cap T^{-n}B$ specifies an $n$-block name followed by an $m$-block name. These sets cover $X$, and there are at most $|\mathcal A||\mathcal B|$ of them. Hence
\begin{align*}
N(\mathcal U_0^{n+m-1})\leq N(\mathcal U_0^{n-1})N(\mathcal U_0^{m-1}).
\end{align*}
After applying $\log$, this is exactly
\begin{align*}
a_{n+m}\leq a_n+a_m.
\end{align*}
[/guided]
[/step]
[step:Apply Fekete's lemma]
The sequence $(a_n)_{n\geq 1}$ is finite-valued and subadditive. By Fekete's lemma,
\begin{align*}
\lim_{n\to\infty}\frac{a_n}{n}
\end{align*}
exists and equals
\begin{align*}
\inf_{n\geq 1}\frac{a_n}{n}.
\end{align*}
Substituting the definition
\begin{align*}
a_n=\log N(\mathcal U_0^{n-1})
\end{align*}
gives
\begin{align*}
\lim_{n\to\infty}\frac{1}{n}\log N(\mathcal U_0^{n-1})
=\inf_{n\geq 1}\frac{1}{n}\log N(\mathcal U_0^{n-1}).
\end{align*}
This proves that the cover entropy limit exists and has the asserted value.
[guided]
Fekete's lemma is the standard existence theorem for subadditive sequences. We have shown that
\begin{align*}
a_{n+m}\leq a_n+a_m
\end{align*}
for all $n,m\geq 1$, and every $a_n$ is finite. Therefore
\begin{align*}
\lim_{n\to\infty}\frac{a_n}{n}=\inf_{n\geq 1}\frac{a_n}{n}.
\end{align*}
Replacing $a_n$ by $\log N(\mathcal U_0^{n-1})$ gives the desired formula for $h_{\mathrm{top}}(T,\mathcal U)$.
[/guided]
[/step]
