Let $X$ be a topological space and $\mathcal{U}$ be an open cover of $X$.
Define $st(M)=\{ U\in\mathcal{U}\vert M\cap U \neq \emptyset \}$ where $M$ is an arbitrary subset of $X$. A cover $\mathcal{V}$ of $X$ is said to be strongly star refinement of $\mathcal{U}$ if $\forall V \in\mathcal{V}$ there exists $U\in\mathcal{U}$ such that $st(V)\subseteq U.$ A cover $\mathcal{U}$ is said to be normal if there exists a sequence $\{\mathcal{U}_n \}_{n\in\mathbb{N}}$ of covers of $X$ such that $\mathcal{U}_1=\mathcal{U}$ and for each $n\in\mathbb{N},$ $\mathcal{U}_{n+1}$ is strongly star refinement of $\mathcal{U}_n.$
It is known that every locally finite cover of a normal space is normal.
My questions are: what type of space guarantees the existence of a normal cover? Does normalness of the space guarantees it, as the name of the cover indicates?
According to this survey paper from the Encylopedia of General Topology, any $T_{3.5}$ space already has some normal covers (the uniform covers form a uniformity that generates its topology. This is clear from the uniform cover axiomatisation. In a paracompact Hausdorff space all open covers are normal (which is nicer to handle).
So yes if your normal includes $T_1$ too (or is "really" $T_4$).