Question: Let $W\_j\subseteq L²(ℝ)$ also Hilbertspaces. If I would like to show
$$L²(ℝ)=\overline{\oplus\_{j\in ℤ}W\_j},$$
can I instead show that 
$$L²(ℝ)=\overline{span\left(\bigcup\limits\_{j\in \mathbb{Z}}^{}W\_j\right)}?$$

Answer: 
In fact
$$\begin{cases}
W\_1 \cap W\_2 = \{ 0 \} \\
(W\_1+ W\_2)\cap W\_3 = \{0\}
\end{cases}$$
are sufficient conditions. The second one implies $(W\_1+W\_2) + W\_3= (W\_1+W\_2) \oplus W\_3$ and the first one that $W\_1 + W\_2 = W\_1\oplus W\_2$. Therefore
$$V= W\_1+W\_2+W\_3 = W\_1\oplus W\_2 \oplus W\_3.$$

However for the general case, I would rather proceed by induction on the dimension $n = \dim V$:

1. Easy for $n=2$...
2. Suppose that the result is true for $n \ge 2$ and that $V = W\_1 +···+ W\_k$ with $\dim V = \displaystyle\sum\_{i=1}^{k} \dim W\_i = n+1$. Without loss of generality, we can suppose that $\dim W\_k \ge 1$. Then $V = (W\_1 +···+ W\_{k-1})+ W\_k$ and $\dim V = \dim (W\_1 +···+ W\_{k-1})+ \dim W\_k$. With what you proved in your question, you have $V = W \oplus W\_k$ where $W = W\_1 +···+ W\_{k-1}$. Now $\dim W \le n$. So you can apply induction hypothesis to it, getting $W = W\_1 \oplus···\oplus W\_{k-1}$ and finally $V = W\_1 \oplus···\oplus W\_{k-1} \oplus W\_k$ as desired.
You have a map $T$ from the external direct sum $W\_1\oplus\cdots W\_k$
to $V$ taking $(w\_1,\ldots,w\_k)$ to $w\_1+\cdots+w\_k$. Its image
is $W\_1+\cdots+W\_k=V$. But both $W\_1\oplus\cdots W\_k$ and $V$ have the same
dimension, so $T$ is an isomorphism. The injectivity of $T$ implies
that inside $V$, $W\_1+\cdots+W\_k$ is an internal direct sum.