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.

ADDED: To fill in details of the argument I've automatically tended to think in terms of Zariski-density and Zariski-closure, but something else must be going on here to deal with the metric topology. This is the point at which I'm doubtful about the strategy used by Kac. But given the brevity of the argument it's probably necessary to look further into the surrounding material for some kind of insight. (Maybe it's just a question of pointing to the fact that both $X$ and $X'$ are *cones*? It would be easier if the author of the book told us what he was thinking about.)

A. (The finite and affine cases, which come up more often in applications, are a little more straightforward.) Keep in mind that the Tits coneCis defined in general by non-strict inequalities (so it serves as a fundamental domain). But the presence of imaginary roots complicates life, so you have to refer back to earlier sections and assemble steps in (c) very carefully. Probably "mistake" is too strong, but it's not easy to follow. $\endgroup$Cthe "Tits cone", but that label belongs to theW-saturated setX. Definite "mistake".] SinceX'is closed and is shown to containX, it also contains the closure ofX. The proof isn't written down carefully, but I think if you assemble the various steps they do add up to a correct proof of the reverse inclusion for the closure ofX. ["Misprint"?] This book is known for its rough spots, even the 3rd edition, but it's an essential source. The only similar textbooks are those by Moody-Pianzola and Carter, which have different emphases. Good luck $\endgroup$