Show that two subspaces $X$ and $Y$ of $\ell^1 (\mathbb{R})$ are such that $\overline{X+Y} = \ell^1 (\mathbb{R})$.

Let $E = \left(\ell^1 (\mathbb{R}),\lVert \cdot\rVert_1\right)$ and consider the subspaces

$$X = \left\{\left(x_n\right)_{n\in \mathbb{Z}_{>0}} \in E: x_{2n} = 0, \forall n\geq 1\right\},\hspace{1em} Y = \left\{\left(y_n\right)_{n\in \mathbb{Z}_{>0}} \in E: y_{2n} = \frac{1}{2^{n+1}}\, y_{2n-1}, \forall n\geq 1\right\}.$$

I'd like to show that $\overline{X+Y} = E$. I've already shown that $X$ and $Y$ are individually closed and that $X+Y\neq E$.

My approach was to show that $X+Y$ has non-empty intersection with all possible open balls $B_{\varepsilon} (z)$, $z\in E$ and $\varepsilon>0$. However, I haven't been able to come up with a sequence $x+y\in X+Y$ such that $\lVert x+y-z\rVert_1 < \varepsilon$ for any $\varepsilon>0$.

Help would be appreciated.