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$C_2^2$ is disjointly* covered by $3$ proper subgroups. I couldn't find any other group that is disjointly covered by $3$ proper subgroups. The groups $C_2\times C_{2n}$ for $n\geq 0$ (i define $C_0 = \mathbb Z$) are all covered by $3$ proper subgroups, but not disjointly so unless $n=1$. This can be verified by looking at the subgroups generated by $(1,0), (0,1)$ and $(1,1)$, and it will be a disjoint cover iff $n=1$.

Are there other (finite?) groups which are (disjointly?) covered by $3$ proper subgroups? Has there been a classification of them?

*by disjoint i mean up to the identity element

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    $\begingroup$ There is a classification: the group must have $C_2\times C_2$ as a quotient, and this is sufficient. I'll dig up a reference later today. $\endgroup$
    – Steve D
    Commented Aug 2 at 19:48
  • $\begingroup$ What do you mean by "disjointly?" Subgroups must all include $1,$ so they can't be disjoint. $\endgroup$
    – Thomas Andrews
    Commented Aug 2 at 21:09
  • $\begingroup$ @ThomasAndrews See the final line in my question.. $\endgroup$
    – Carla_
    Commented Aug 2 at 21:33

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This is answered in Bhargava's paper Groups as Unions of Proper Subgroups. There it is shown that a group is the union of $3$ proper subgroups if and only if it has $C_2\times C_2$ as a quotient. No finiteness assumption is necessary.

The proof shows that this union is "disjoint" precisely when the kernel is trivial, so the group is isomorphic to $C_2\times C_2$.

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    $\begingroup$ Bhargava's paper, which appears in The American Mathematical Monthly, May 2009, attributes this theorem to Gaetano Scorza. $\endgroup$
    – Lee Mosher
    Commented Aug 2 at 20:23

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