After taking a closer look at the proof by Kac of Prop. 5.8 c), I can see that it's too sketchy to be followed easily. Here the generalized Cartan matrix is assumed to be of indefinite type, which I have little experience with. But the basic steps in the proof might be organized as follows:
In the background is the infinite root system $\Delta= \Delta_+ \cup \Delta_-$ along with a further partition $\Delta_+ = \Delta_+^{re} \cup \Delta_+^{im}$. Here $\Delta_+$ consists of $\mathbb{Z}^+$-linear combinations of the fixed simple roots $\alpha_1, \dots, \alpha_n$. The $\alpha_i$ are linearly independent and lie in the dual space of $\mathfrak{h}_\mathbb{R}$ where $\mathfrak{h}$ is the finite dimensional Cartan subalgebra. The Weyl group $W$ is generated by all reflections $r_\alpha$ with $\alpha \in \Delta_+$.
The Tits cone $X$ is the image under $W$ of $C:= \{ h \in \mathfrak{h}_\mathbb{R} | \langle \alpha_i, h \rangle \geq 0 \text{ for all } i\}$. The candidate for its metric closure is $X' := \{h \in \mathfrak{h}_\mathbb{R} | \langle \beta, h \rangle \geq 0 \text{ for all } \beta \in \Delta_+^{im}\}$. Being defined by inequalities, $X'$ is closed.
By Prop. 5.2 a), $\Delta_+^{im}$ is $W$-invariant (so $X'$ is). Obviously $X' \supset C $, so $X' \supset \overline{X}$.
In the reverse direction, consider just those $h \in X'$ for which $\langle \alpha_i, h \rangle \in \mathbb{Z}$ for all $i$. These elements of $X'$ are dense in the metric topology (just as $\mathbb{Z}$ is dense in $\mathbb{R}$), so it's enough to show they all lie in $X$ (where they will form a dense subset of $\overline{X}$).
Use Thm. 5.6 c) to find $\beta \in \Delta_+^{im}$ such that the "Cartan integers" $\langle \beta, \alpha_i^\vee \rangle <0$ for all $i$. (So $\beta= \sum_i b_i \alpha_i$ with all $b_i>0$.)
In turn, for all $\gamma \in \Delta_+^{re}$, $$r_\gamma (\beta) = \beta - \langle \beta, \gamma^\vee \rangle \gamma = \beta + s \gamma$$ with $s$ larger than the sum of coefficients of $\gamma$ because $\langle \beta, \alpha_i^\vee \rangle <0$ for all $i$. Thanks to Prop. 5.2 c), all $\beta +s \gamma \in \Delta_+^{im}$. Since $h \in X'$, it follows that $\langle \beta + s\gamma, h \rangle \in \mathbb{Z}^+$. In particular, only finitely many such $\gamma$ exist with $\langle \gamma, h \rangle \leq -1$.
But Prop. 3.12 c) characterizes $X$ as the set of all $h$ for which only finitely many $\gamma \in \Delta_+$ satisfy $\langle \gamma, h \rangle <0$. Combined with the special choice of $h$, we get $h \in X$ as desired.