Here's basically the same answer as Mikko Korhonen, but written in a way that's slightly easier for me to understand.

Let me use $\leq$ to denote the partial order on all of the vector space $E$ with $u \leq v$ for $u,v \in E$ if and only if $v-u = \sum_{i=1}^{n}a_i \alpha_i$ with all $a_i \geq 0$ (but not necessarily integral). Considering $\leq$ instead of $\preceq$ means I don't have to deal with the different classes of the root lattice mod the weight lattice.

Fix any $v \in E$. We claim there are only finitely many dominant weights $\mu \in \Lambda^{+}$ for which we don't have $v \leq \mu$, from which the desired claim obviously follows. Indeed, let $\omega$ be a fundamental weight. As Mikko explains (and as is also fundamental to the answer I gave to the linked question), we have $\omega = \sum_{i=1}^{n} a_i \alpha_i$ with all $a_i > 0$. Hence clearly there is some $m \geq 0$ so that $v \leq m\omega$. Now let $M$ be the maximum over all fundamental weights of such $m$. Then writing $\mu = \sum_{i=1}^{n} c_i \omega_i$, the only way we could fail to have $v \leq \mu$ is if $c_i < M$ for all $i$. (Here we are using the fact that if $v \leq u$ and $\nu \in \Lambda^{+}$, then $v \leq u+\nu$, which can be seen again from the fact that writing $\omega = \sum_{i=1}^{n} a_i \alpha_i$ for any fundamental weight $\omega$, we have $a_i > 0$.) There are clearly only finitely many $\mu=\sum_{i=1}^{n} c_i \omega_i \in \Lambda^{+}$ with $c_i < M$ for all $i$.

EDIT:

Here is an even more general statement/context. Let $V$ be an n-dimensional Euclidean vector space with inner product $\langle \cdot, \cdot \rangle$ and let $v_1,\ldots,v_n \in V$ be a collection of vectors such that:

- $v_1,\ldots,v_n$ form a basis of $V$;
- $\langle v_i , v_j \rangle \leq 0$ for all $i \neq j$ (in other words, the vectors are pairwise non-acute);
- there is no nontrivial decomposition of $V = V_1 \oplus V_2$ into orthogonal subspaces $V_1$ and $V_2$ such that $v_i \in V_1 \cup V_2$ for all $i$ (this is an irreducibility condition- equivalently it says that if we draw the graph on the $v_i$ with $v_i$ adjacent to $v_j$ if $\langle v_i, v_j\rangle < 0$ then that graph will be connected).

Then let $Q_{\geq 0} := \{ v=\sum_{i=1}^{n}a_iv_i, a_i \geq 0\}$ be the cone generated by the $v_i$. And let $P_{\geq 0} := \{v \in V\colon \langle v,w\rangle \geq 0 \textrm{ for all $w\in Q_{\geq 0}$}\}$ be the dual cone to $Q_{\geq 0}$.

Then the claim is that for any $v\in V$ we have that $P_{\geq 0} \setminus (v+Q_{\geq 0})$ is a bounded subset of $V$.

To prove this, observe that any nonzero $w \in P_{\geq 0}$ (in particular, any generator of this cone) has all $a_i > 0$ when we write $w=\sum_{i=1}^{n} a_i v_i$. The reason for this is that the matrix $M=(\langle v_i, v_j \rangle)$ is a nonsingular, irreducible $M$-matrix, and hence $M^{-1}$ (which expresses the coordinates of the generators of $P_{\geq 0}$) is a matrix with all entries strictly positive (see e.g. Theorem A of https://core.ac.uk/download/pdf/82640451.pdf). Then we can apply the same argument as above to the generators of the cone $P_{\geq 0}$.

(The particular situation above corresponds to the $v_i$ being the simple roots and the cone $P_{\geq 0}$ being the dominant cone, i.e., cone spanned by the fundamental weights.)