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, 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.