# Open set minus closed set with empty interior

Let $$X$$ be a separable infinite-dimensional Banach space, $$U$$ be non-empty and open in $$X$$ and $$E$$ be finite-dimensional linear subspace of $$X$$.

I would like to know if there is a non-empty open subset of $$U$$ which does not intersect $$E$$

What I've done so far:

Conjecture: Yes

Since $$X$$ is infinite dimensional and $$E$$ is finite-dimensional, then $$E$$ must have empty interior (since no ball can be contained in $$E$$). Since $$E$$ has an empty interior and every open subset $$V\subseteq U$$ is open, then $$E$$ cannot contain $$U$$ or any of its open subsets $$V$$. Hence, $$V \not\subseteq E \qquad (\forall V\subseteq U,\, \mbox{V open})$$

Let us show that there is a non-empty open subset $$U^*\subseteq U$$ which does not intersect $$E$$.

In particular, let $$Ball(x,r)$$ be an open ball in $$U$$ of radius $$r>0$$ centered at some $$x \in U$$. By translation, $$Ball(0,r)$$ and $$E-x$$ are both of the same for as above, expect now $$Ball(0,r)$$ contains a countable basis $$\{x_n\}_{n \in \mathbb{N}}$$ (by separability) of $$X$$.

In particular, there are $$\{r_n>0\}_{n \in \mathbb{R}}$$ such that the open balls $$\{Ball(x_n,r_n)\}_{n \in \mathbb{N}}$$ are disjoint, contained within $$Ball(0,r)$$, and the vectors in $$Ball(x_n,r_n)$$ are linearly independent from $$span\{x_i:i=1,\dots,n\}$$. (I assume this should be possibly by inductively choosing a small enough radius)

If $$E-x$$ intersected each of these then it would contain a basis for $$X$$, contradicting its finite-dimensionality. Therefore, there is some $$N\in\mathbb{N}$$ for which $$E-x \cap Ball(x_n,r_n)=\emptyset$$; whence $$E \cap Ball(x_n+x,r_n)=\emptyset.$$

Comments: I feel that the existence of the open balls was a bit unconvincing and likewise for the fact that $$E$$ must contain a basis since is does intersect all of those balls.

An open set $$U$$ minus a closed set $$C$$, regardless of which topology, is always open. You are taking the intersection of an open set, and the complement of a closed set, which is open, and thus the resulting set $$U \setminus C$$ is open.
The only question is, can the resulting set be empty? The empty set is a perfectly valid open set. Note that $$U \setminus C = \emptyset$$ if and only if $$U \subseteq C$$, which is to say, either $$U = \emptyset$$ or $$C \subseteq \operatorname{int} U \neq \emptyset$$.
As you've pointed out, finite-dimensional subspaces of infinite-dimensional spaces are closed and have empty interior. Thus, we cannot have $$U \setminus C = \emptyset$$; thus $$U \setminus C$$ is the open subset of $$U$$, disjoint from $$C$$, that you're looking for.
Since $$E$$ is finite-dimensional, it is a closed subset of $$X$$ and therefore you can just take $$U=E^\complement$$.