With respect to collisions, hashing twice can not increase security, because if $x$ and $x'\ne x$ collide for $H$, that is $H(x)=H(x')$, then $H(H(x))=H(H(x'))$. Otherwise said, any collision for $H$ is a collision for the double hash $H\circ H$. It is therefore trivial to exhibit collisions for $\operatorname{MD5}\circ\operatorname{MD5}$. Hence the answer to the question as worded in its title is NO.
With respect to preimage, hashing twice demonstrably does not harm security (from a preimage for $H\circ H$ one can make a preimage for $H$, simply by applying $H$), and tends to improve it for practical functions. In particular, one hypothetically able to build preimages for $\operatorname{MD5}$ only for 512-bit messages would likely have a hard time extending that to $\operatorname{MD5}\circ\operatorname{MD5}$.
Handwaving argument: if for some 128-bit $v$ one could find a 512-bit $\operatorname{MD5}\circ\operatorname{MD5}$ premiage $m$, then the easilly computed $m'=\operatorname{MD5}(m)$ would be a 128-bit $\operatorname{MD5}$ preimage of $v$, which arguably is harder to find than a 512-bit $\operatorname{MD5}$ preimage of $v$, since for a given $v$ it is expected that there are about $1$ premimage of the former kind, and about $2^{384}$ of the later.
One area where double-hashing increases security is length extension attack. $\operatorname{SHA-256}$ is trivially vulnerable to that, $\operatorname{SHA-256}\circ\operatorname{SHA-256}$ is not.
In some use cases, hashing twice can destroy security, including for common hashes; an example is given here; in summary: some proof-of-work protocol safe when using $H$ is entirely unsafe using $H\circ H$.
With usual hash functions (or ideal ones), hashing twice does not dramatically reduce the output space (it is reduced by a factor about $1-1/e\approx0.63212$); that makes accidental collision slightly less unlikely, and it is mostly immaterial in practice. It does not make collisions easier to exhibit, since the amount of invocations of $H$ expected necessary to exhibit a collision by whatever brute force method can not decrease (if it did, that would be trivially usable to exhibit collisions for $H$ at reduced cost).
