Given any finite-dimensional normed space, the topology is equivalent to that generated by the Euclidean norm. So the Euclidean open ball is norm closed. The fact that is is strictly convex is independent of the topology. Hence we have the following
Lemma. Every finite dimensional normed space has a strictly convex neighborhood of the origin.
I wonder is this true for infinite dimensional spaces? It seems for Banach spaces a lot is known about conditions for admitting an equivalent strictly convex norm. For example such a norm exists if the space is separable, and there exist counterexamples otherwise.
But I don't need the norm to be equivalent. I would be satisfied with a continuous strictly convex norm. As then the open unit ball would be open wrt the original norm, and would be the set I want.
One can attempt to approach the problem using Hamel bases. Given a Hamel Basis $v_\alpha$ define the new norm
$$| c_1 v_1+c_2v_2+\ldots + c_n v_n| = \sqrt{c_1^2 + \ldots + c_n^2}$$
and this indeed gives a strictly convex norm on the original space. However it is not continuous since there will always be lots of discontinuous Hamel functionals. So the unit ball will not be open wrt the original norm.
Does such a set always exist?