1. Initial Setup: Let $\mathcal{F}$ be a finite collection of open discs in $\mathbb{R}^2$ whose union contains a set $E \subseteq \mathbb{R}^2$. We need to find a pairwise disjoint subcollection $D_1, \ldots, D_n$ in $\mathcal{F}$ such that $E \subseteq \bigcup_{j=1}^n 3D_j$.

2. Selection Process: We will construct the subcollection $D_1, \ldots, D_n$ iteratively. Start by selecting the largest disc $D_1$ from $\mathcal{F}$. Then, for each subsequent disc $D_i$, choose the largest disc from $\mathcal{F}$ that is disjoint from all previously selected discs $D_1, \ldots, D_{i-1}$. This process is possible because $\mathcal{F}$ is finite, and if multiple discs satisfy the condition, we can choose any one of them arbitrarily.

3. Termination: This selection process will terminate after a finite number of steps because $\mathcal{F}$ is finite. Let the resulting subcollection be $D_1, \ldots, D_n$.

4. Proof of Coverage: We need to show that $E \subseteq \bigcup_{j=1}^n 3D_j$. Consider any disc $E_k \in \mathcal{F}$ that is not in the subcollection $D_1, \ldots, D_n$. By construction, $E_k$ must intersect with at least one of the discs in the subcollection $D_1, \ldots, D_n$. Let $D_i$ be one such disc with which $E_k$ intersects.

5. Containment in Enlarged Discs: Since $D_i$ was chosen to be the largest disc that is disjoint from the previously selected discs, it follows that the radius of $D_i$ is at least as large as the radius of $E_k$. Therefore, the disc $3D_i$ (which has three times the radius of $D_i$) will completely contain $E_k$.

6. Conclusion: Since every disc $E_k \in \mathcal{F}$ is either in the subcollection $D_1, \ldots, D_n$ or intersects with one of these discs and is contained in the corresponding $3D_i$, it follows that the union of the enlarged discs $3D_1, \ldots, 3D_n$ will cover the entire set $E$. Hence, $E \subseteq \bigcup_{j=1}^n 3D_j$.

$$\blacksquare$$