Let $K$ be a non-Archimedean local field, and let $\mathcal{O}$ be its ring of integers. Let $U$ be the group of $4 \times 4$ unipotent upper-triangular matrices over $K$. Consider the topology on $K$ induced by its normalized absolute value, and identify $U$ with the product space $K^6$ e.g. by the map
$$ \begin{pmatrix} 1 & u_{12} & u_{13} & u_{14} \\ & 1 & u_{23} & u_{24}\\ & & 1 & u_{34} \\ & & & 1 \end{pmatrix} \mapsto (u_{12},u_{13},u_{14},u_{23},u_{24},u_{34})$$
Then $U$ is a locally profinite group with respect to the topology induced by $K^6$. Now, fix $a \in K$ with $a \neq 0$. Since $\mathcal{O}$ is compact and open in $K$, the set
$$ N:=\left\{ \begin{pmatrix} 1 & 0 & a\alpha & 0 \\ & 1 & \alpha & 0 \\ & & 1 & 0 \\ & & & 1 \end{pmatrix} : \alpha \in \mathcal{O} \right\} $$ is a compact open subgroup of $U$. On the other hand, the subgroup
$$ L:= \begin{pmatrix} 1 & 0 & 0 & * \\ & 1 & * & * \\ & & 1 & * \\ & & & 1 \end{pmatrix} $$
is closed in $U$, so that $L \cap N$ is a compact open subgroup of $L$. But $ L \cap N =\{I_4\}$, which implies that $L$ is discrete. In particular, the "root" subgroup
$$X:= \left\{ \begin{pmatrix} 1 & 0 & 0 & 0 \\ & 1 & \alpha & 0 \\ & & 1 & 0 \\ & & & 1 \end{pmatrix} : \alpha \in \mathcal{O} \right\}$$ is contained in $L$, and so it is also discrete with respect to the subspace topology. As $X$ is homeomorphic to $\mathcal{O}$ and $\mathcal{O}$ is compact, this implies that $\mathcal{O}$ must be finite. But this is absurd (e.g. take the field $K=\mathbf{Q}_p$ of $p-$adic numbers, whose ring of integers $\mathcal{O}$ is the ring $\mathbf{Z}_p$ of $p-$adic integers).
I stumbled upon this by accident, and I'm desperately struggling with finding the error. Thank you in advance.
Sorry maybe I misunderstood your notations, but what is your I_4 ? the identity matrix ? If so you don’t get
$L\intersect N = I_4$