Indecomposable elements in a lattice

Let $I$ be the set of indecomposable elements in $L$.

1. Hopefully you're aware that while the condition

   $u^{\top} v \geqslant 0$ implies $H_u \cap H_v \subseteq H_{u + v}$

   is essential in proving that $V(0) = \bigcap \limits_{a \in I} H_a$, it is far from being sufficient on its own.

2. You're trying to prove that

   • $I$ is a minimal set satisfying $V(0) = \bigcap \limits_{a \in I} H_a$, i.e. for any $b \in I$ we have that $V(0) \subsetneq \bigcap \limits_{a \in I \setminus \{ b \}} H_a$;
   • $I$ is a unique such minimal set.
     
     

   For this it suffices to prove a stronger statement:

   $(*) \quad$ If $A \subseteq L$ is a subset satisfying $V(0) = \bigcap \limits_{a \in A} H_a$, then $I \subseteq A$.

   Be advised: I'm assuming that the inequality $b^{\top} c > 0$ in the definition of indecomposability should be non-strict (i.e. $b^{\top} c \geqslant 0$), as otherwise the statement from the first bullet is false - $I$ need not be minimal. An easy counterexample is $\mathbb{Z}^2 \subseteq \mathbb{R}^2$, where $V(0)$ is generated (by means of intersecting the $H_a$'s) by just four elements: $(1, 0)$, $(0, 1)$, $(-1, 0)$, $(0, -1)$, but $(1, 1)$ is also indecomposable.

   It remains to prove $(*)$. First note that $0 \notin A$ as $H_0 = \varnothing$. Now fix any $b \in I$. Clearly $\frac{1}{2} b \notin V(0)$ as $\frac{1}{2}b \notin H_b$, so there is $a \in A$ such that $\frac{1}{2}b \notin H_a$, i.e. $\| \frac{1}{2} b \| \geqslant \| \frac{1}{2} b - a \|$ or (after easy transformations) $\left< a, b-a \right> \geqslant 0$. But then we may write $b = a + (b-a)$, so by the indecomposability of $b$ one of the summands must be zero, which implies $b = a$ and therefore $b \in A$. $\square$

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3. A proof of $V(0) = \bigcap \limits_{a \in I} H_a$, requested in the comment. I will use the following lemma:

   If $A \subseteq L$ is a nonempty subset, then there is $b \in L$ such that $\| b \| = \min \limits_{a \in L} \| a \|$.

   Proof: assume for contradiction that such $b$ does not exist. Then there is a sequence $(a_n)$ of elements of $A$ such that $\| a_{n+1} \| < \| a_n \|$ for each $n$. Such a sequence must be bounded, so it has a limit point. As $L$ is a subgroup, it follows that $L$ contains points arbitrarily close to $0$, which contradicts the assumption that $L$ is a lattice, so the lemma has been proved.

   Now assume for contradiction that $\bigcap \limits_{a \in I} H_a \neq V(0)$, which means that $\bigcap \limits_{a \in I} H_a \not \subseteq H_b$ for some $b \in L \setminus \{ 0 \}$. By the lemma, we can assume that $b$ has the smallest distance to $0$ of all elements of $L \setminus \{ 0 \}$ with that property, i.e. $\bigcap \limits_{a \in I} H_a \subseteq H_c$ for every $c \in L \setminus \{ 0 \}$ with $\| c \| < \| b \|$.

   Now clearly $b \notin I$, so write $b = c+d$ where $c, d \neq 0$ and $\left< c, d \right> \geqslant 0$. Then $\| b \|^2 = \| c \|^2 + \| d \|^2 + 2 \left< c, d \right> \geqslant \| c \|^2 + \| d \|^2$, thus $\| c \|, \| d \| < \| b \|$. By the choice of $b$ we have that $\bigcap \limits_{a \in I} H_a \subseteq H_c$ and $\bigcap \limits_{a \in I} H_a \subseteq H_d$. But $H_c \cap H_d \subseteq H_{c+d} = H_b$, which is a contradiction.