This is an extended comment Qiaochu answer; so extended indeed that it is larger than its progenitor. (Warning: I largely made it up on the go, so there may be mistakes.)
First, the concrete. Consider the "mixed Heisenberg group": namely, the set of "matrices"
$\begin{pmatrix}
1 & a & c\\
0 & 1 & b\\
0 & 0 & 1\\
\end{pmatrix}$
with $a, b \in \mathbb{Z}$ and $c \in \mathbb{R}$. Since $\mathbb{Z} \subset \mathbb{R}$ we can consider this a subset of the real Heisenberg group. The formulas
$\begin{pmatrix}
1 & a & c\\
0 & 1 & b\\
0 & 0 & 1\\
\end{pmatrix}
\begin{pmatrix}
1 & a' & c'\\
0 & 1 & b'\\
0 & 0 & 1\\
\end{pmatrix}=
\begin{pmatrix}
1 & a+a' & c+c'+ab'\\
0 & 1 & b+b'\\
0 & 0 & 1\\
\end{pmatrix}$ and
$\begin{pmatrix}
1 & a & c\\
0 & 1 & b\\
0 & 0 & 1\\
\end{pmatrix}^{-1}=
\begin{pmatrix}
1 & -a & ab-c\\
0 & 1 & -b\\
0 & 0 & 1\\
\end{pmatrix}$ convince us that this is a subgroup.
Now this is pretty clearly a Lie subgroup (it's a closed submanifold; you can check things explicitly or apply Cartan's theorem (an overkill in this case)).
It is a central extension of $\mathbb{Z}^2$ by $\mathbb{R}$ and is not a direct product, as in Quiaochu's answer.
Let's spell out the check: the element of $H^2(\pi_0(G), G_0)$ Qiaochu referred to has a representative constructed by choosing a set-theoretic section $s$ (in this case there is an obvious one, sending $(a, b)$ to $\begin{pmatrix}
1 & a & 0\\
0 & 1 & b\\
0 & 0 & 1\\
\end{pmatrix}$), and measuring its failure to be a group homomorphism
$\phi((a,b), (a', b'))=s((a,b)) s((a',b'))s((a+a', b+b'))^{-1} \in G_0=\mathbb{R}$
In our case this is equal to $ab'$. One can fairly easily check that this $\phi$ is not $dh((a,b), (a', b'))=h((a,b))+h((a',b')-h((a+a',b+b'))$ for any $h:\mathbb{Z}^2 \to \mathbb{R}$ (plug in $a=b'=0$ and then $a'=b=0$ and compare the results), hence the class is non-trivial, so the extension is not split (the map $\phi$ changes when we change a section $s$ to $t$ by exactly $d(st^{-1})$ where $(st^{-1})$ is viewed as map $\mathbb{Z}^2 \to \mathbb{R}$; thus the class is well defined - this is what one checks to see that $H^2(Q, N)$ classifies extensions).
We can also take a "mixed Heisenberg group mod n" for an odd prime, meaning that $a,b \in \mathbb{Z}/n\mathbb{Z}$ and $c \in \mathbb{R}/n\mathbb{Z}$. The above construction still works, but everything is now compact/finite.
Now, the general. I claim that in general the construction described by Qiaochu naturally yields a Lie group - because the extension is central. Let $p: G\to \pi_0(G)$ be the projection map. Topologically, define a subset $U$ of $G$ contained in $G_b=p^{-1}(b)$ to be open if for some $y \in G_b$ the set $y^{-1}U$ is open. [Notes: 1) $x y^{-1}= y^{-1}(xy^{-1})y=y^{-1}x$, so it does not matter which side we multiply on 2) if $z \in G_b$ then $z^{-1}U= (z^{-1}y)y^{-1}U$ is open if and only if $y^{-1}U$ is open, so it does not matter which $y \in G_b$ we use.] Now declare any subset of $G$ open if its intersection with each $G_b$ is open.
Around any $g\in G$ one gets a chart by composing left (or right, these are the same) multiplication by $g^{-1}$ with a chart $C: U_e \to \mathbb{R}^n$ in $G_0$ around identity. Now we need to check that inverses and multiplication are smooth; however, because $G_0$ is in the center, the maps on subsets in $\mathbb{R}^n$ induced via the above charts are equal to the maps for corresponding maps in $G_0$ (for example, using $inv$ to denote map inverse and ${-1}$ to denote group inverse, we have $C( g[(g C^{inv}(r))^{-1}])=C((C^{inv}(r))^{-1})$; etc.) So $G$ is indeed a Lie group.
Now if the extension is not central presumably bad things may happen if action of $Q$ on $N$ is not via Lie automorphisms (this seems fairly "easy" to imagine, just take any Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$ and define additive map $f:\mathbb{R}\mapsto\mathbb{R}$ by permuting the basis vectors by arbitrary non-trivial permutation, thus $f \in Aut(\mathbb{R})$; now define semidirect product of $\mathbb{R}$ and $\mathbb{Z}$ using this (you can even take a permutation of order 2 and use $\mathbb{Z}/2\mathbb{Z}$ instead of $\mathbb{Z}$...). However, if the action is by $Aut(N)$ then the semidirect product is also a Lie group. It seems that for non-split extensions with such an action one should also get a Lie group structure on $G$, but I have not checked the details.
