Suppose $\mathcal T$ and $\mathcal S$ are two compatible Hausdorff topologies on a group $G$ and $\mathcal R$ is a maximum compatible topology on $G$ with $\mathcal R \subseteq \mathcal T\cap \mathcal S$.
Is $(G,\mathcal R)$ Hausdorff?
Suppose $\mathcal T$ and $\mathcal S$ are two compatible Hausdorff topologies on a group $G$ and $\mathcal R$ is a maximum compatible topology on $G$ with $\mathcal R \subseteq \mathcal T\cap \mathcal S$. Is $(G,\mathcal R)$ Hausdorff? 


It seems the answer is no. Basically, the $p$adic and Archimedean topologies on $\mathbb{Q}$ are incompatible enough that the maximal compatible topology contained in both of them is indiscrete. Here is an outline of the proof (I'm heading to bed so I haven't filled in the details): $\bullet$ Any subset of $\mathbb{Q}$ which is open for both topologies must be dense for both topologies. Therefore, all elements of $\mathcal{R}$ are dense. $\bullet$ If $\mathcal{R}$ is compatible with the group structure, the complement of the antidiagonal in $\mathbb{Q}\times \mathbb{Q}$ must be open. Since this complement contains e.g. $(1,1)$, it must contain a set of the form $U\times U$ for some $U\in \mathcal{R}$. $\bullet$ This is impossible since $U\cap U^{1}\neq \emptyset$ by density of $U$. 


Edit: This answer is completely wrong. Please ignore it, and see Kevin's answer. A $T_{0}$topological group $G$ is Hausdorff ($T_{2}$). Recall that $T_{0}$ means that for all $x,y \in G$, there is an open subset $U$ such that $x \in U, y \notin U$ or $y \in U, x \notin U$. Consequently (by homogeneity), a topological group is Hausdorff if and only if the one element subgroup of the unit element is closed. Thus, for your question it suffices to show that $\{1\} \subset G$ is closed for the $\mathcal{R}$topology. This is clear if we show that $\mathcal{R} = \mathcal{T} \cap \mathcal{S}$. Let $U$ be an open subset for $\mathcal{T}$ and $\mathcal{S}$. For every element $g \in G$, the set $gU$ is open for $\mathcal{T}$ and $\mathcal{S}$, because both were compatible. For the same reason $U^{1}$ is open for both $\mathcal{T}$ and $\mathcal{S}$. Hence $\mathcal{R} = \mathcal{T} \cap \mathcal{S}$ is compatible. Consequently $U = G  \{1\}$ is an open subset for $\mathcal{R}$, and hence $(G,\mathcal{R})$ is Hausdorff. 

