Haar-null union of dense subsets Let $\{X_i\}_{i \in \mathbb{R}-\{0\}}$ be a set of subsets of a separable infinite-dimensional Fréchet space $X$ and $I$ be uncountable.  Moreover, suppose that 


*

*(Dense $G_{\delta}$) $X_i$ is a dense $G_{\delta}$ subset of $X$ not containing $0$,

*(Almost Contains a Linear Subspace) For each $i$, there exists a dense linear subset $E_i\subset X$ satisfying
$$
E_i-\{0\}\subseteq X_i
$$

*(Disjoint) $\bigcap_{i \in I} X_i=\emptyset$,

*(Not a Cover) $\cup_{i \in I} X_i \neq X-\{0\}$,


Can we conclude that:
$$
X - \bigcup_{i \in \mathbb{R}-\{0\}} X_i,
$$
is Haar-null, or at-least it is finite-dimensional?  
I have never seen this type of result and am pretty new to this type of thing but I ask here since it seems beyond the level of math-stack exchange. 
Relevant Definitions:
Haar-null set:  A subset $A\subseteq X$ is Haar-null if there exists a Borel probability measure $\mu$ on $X$ and a Borel subset $A\subseteq B$ satisfying
$$
\mu\left(
B+x
\right)=0 \qquad (\forall x \in X).
$$

Facts:


*

*I do know that $X=X_i -X_i$ upon applying the Baire category theorem.  (Also from the comments the Pettis Lemma).  This means that every element in $X$ can be represented as a sum of elements from each $X_i$.

*In the case (not covered by my question) where $I$ is a singleton, this paper gives a counter-example.



Intuitions:
As intuition, it can be seen here, that if $X$ is locally compact, then a Borel set is Haar-null if and only if it is of Haar-measure $0$.  
 A: In the Frechet space $X:=\mathbb R^\omega$ consider the dense linear subspace $$L_0:=\{(x_n)_{n\in\omega}\in\mathbb R^\omega:|\{n\in\omega:x_n\ne0\}|<\omega\}.$$
Fix a countable base $\{V_n\}_{n\in\omega}$ of the topology of the space $L_0$ and in each set $V_n$ choose a point $x_n$, which is not contained in the linear hull of the set $\{x_i\}_{i<n}$. Then $\{x_n\}_{n\in\omega}$ is a dense linearly independent set $\{x_n\}_{n\in\omega}$ in $X$. For every $n\in\mathbb N$ consider the linear hull $L_n$ of the set $\{x_m\}_{m\ge n}$ and observe that $\{x_m\}_{m\ge n}$ and $L_n$ are dense in $X$, and $\bigcap_{n\in\omega}L_n=\{0\}$.
Consequently, for every non-zero element $x\in X$ we can find a number $n_x\in \omega$ such that $x\notin L_{n_x}$. 
It is easy to see that the closed convex set $F:=[1,\infty)^\omega$ in $X=\mathbb R^\omega$ is not Haar-null but is disjoint with the dense linear subspace $L_0$ of $X$.
For any $x\in X\setminus\{0\}$ consider the open subset $W_x:=X\setminus(F\cup \cup\{x,0\})$ and observe that $L_{n_x}\setminus\{0\}\subset W_x\subset X\setminus\{x,0\}$, which implies $\bigcap_{x\in X\setminus \{0\}}W_x=\emptyset$.
Also $X\setminus \bigcup_{x\in X\setminus\{0\}}W_x\supset F$ is not Haar-null.
So, the family of dense open (and hence $G_\delta$) sets $(W_x)_{x\in X\setminus\{x\}}$ has the properties required in the question.
