Open sets with the required property are exactly functionally open sets.
Let us recall that a subset $U$ of a topological space $X$ is functionally open in $X$ if $U=f^{-1}[V]$ for some continuous function $f:X\to \mathbb R$ and some open set $V\subseteq U$$V\subseteq \mathbb R$. It is well-known that an open subset $U$ of a normal space $X$ is functionally open if and only if $U$ is of type $F_\sigma$.
Theorem. For a subset $U$ of a topological space $X$, the following conditions are equivalent:
$U$ is functionally open in $X$;
For any Banach space $Y\ne\{0\}$ there exists a continuous function $f:X\to Y$ such that $f^{-1}[B]=Y$ where $B=\{y\in Y:\|y\|<1\}$ is an open unit ball in $Y$.
Proof The implication (2)$\Rightarrow$(1) is trivial becase the real line is a Banach space. To prove the implication (1)$\Rightarrow$(2), assume that the set $U$ is functionally open in $X$. Then $U=g^{-1}[V]$ for some continuous function $g:X\to\mathbb R$ and some open set $V\subseteq \mathbb R$. If $V=\mathbb R$, then $U=X$ and then $U=X=f^{-1}[B]$ for the constant function $f:X\to \{0\}\subset Y$.
So, we assume that $V\ne \mathbb R$. In this case, consider the continuous function $h:\mathbb R\to[0,1]$, $h:t\mapsto \min\{1,\min_{s\in \mathbb R\setminus V}|t-s|\}$. Observe that $h^{-1}(0)=\mathbb R\setminus V$. Choose any point $y_1\in Y$ with $\|y_1\|=1$ and consider the function $f:X\to Y$, $f:x\mapsto (1-h\circ g(x))y_1$. It is easy to check $f^{-1}[B]=U$. $\square$
