Showing that there is a point belonging to $2000$ sets.

As pointed out by @mathmandan, you have never shown that $f^{-1}([2000,\infty])$ is non-empty, which is equivalent to what the question is asking about.

• For the proof of the first part, let $f = \sum_{i\geq 1} \chi_{E_i}$ and note that $$ \int_{[0,1]} f^2 \, \mathrm{d}m = \sum_{i,j \geq 1} \int_{[0,1]} \chi_{E_i}\chi_{E_j} \, \mathrm{d}m = \sum_{i,j \geq 1} m(E_i \cap E_j) \geq \sum_{i,j \geq 1} \frac{1}{i^2+j^2} = \infty. $$ In particular, $f$ cannot be bounded above and hence the desired claim follows.

• On the other hand, if we set $E_i = (0,i^{-2})$, then $$ m(E_i \cap E_j) = \frac{1}{\max\{i^2, j^2\}} \geq \frac{1}{i^2+j^2}, $$ whereas each $x \in [0, 1]$ lies in only finitely many of $E_i$'s. (In fact, for this choice, we have $\sum_{i\geq 1} \chi_{E_i}(x) = \lfloor\frac{1}{\sqrt{x}}\rfloor$ for any $x \in (0, 1]$.)

I think that what you have said so far is fine...but I don't think you have shown that $f^{-1}([2000, \infty])$ is nonempty.

Consider the following sequence: $$ E_i = \left(\frac{1}{2^{i}}, \frac{1}{2^{i-1}}\right) $$

So $E_1 = \left(\frac{1}{2}, 1\right)$, and $E_2 = \left(\frac{1}{4}, \frac{1}{2}\right)$, etc. No point belongs to more than one of these sets; they are pairwise disjoint.

So there is no point which belongs to $2000$ or more of these sets. In other words, $f^{-1}([2000, \infty]) = \emptyset$. (This example shows that you'll need to use the given condition somehow.)