Why is the string group not a Lie group? The string group $String(n)$ is by definition a 3-connected cover of $Spin(n)$. This definition determines the homotopy type of the string group.
[In a previous version of this question I screwed up the definition and caused some confusion, see the comments below.]
A common argument is saying that "the string group cannot be a Lie group because it has vanishing $\pi_3$". This is obviously not a complete argument because $(\mathbb{R},+)$ is a nice Lie group with vanishing $\pi_3$. 
What is the correct statement about Lie group structures on the string group, and how does one prove it?
 A: To follow up, there is now an infinite-dimensional Lie group model of String:

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*Thomas Nikolaus, Christoph Sachse, Christoph Wockel, A Smooth Model for the String Group, Int. Math. Res. Not. 16 (2013) 3678-3721, doi:10.1093/imrn/rns154, arXiv:1104.4288.

A: As David Roberts is saying it's conceivable the string group could be represented by an infinite dimension manifold. I'm totally agnostic on that, but as I interpret the question it's asking why it's not equivalent (as an H-space?) to a non-compact finite dimensional Lie group (David Roberts also explains that for a compact simply connected Lie group we always have $\pi_3$ non vanishing). I think though the underlying space has cohomology in infinitely many dimensions. Let me illustrate this in the case of $\mathrm{String}(3)$. So we have a Serre spectral sequence for the fibration $K(\mathbb{Z},2)\to \mathrm{String}(3) \to S^3$. Now thinking of $\mathbb{Z}[x]$ as the cohomology ring of $K(\mathbb{Z},2)$, the differential has to be $d:x \mapsto e$, the generator for the cohomology of $S^3$. So using the Leibnitz rule, $x^2\mapsto 2x\otimes e$,  $x^3 \mapsto 3x^2\otimes e$... etc. This means that $H^5(\mathrm{String}(3))= \mathbb{Z}/2\mathbb{Z}$, $H^7(\mathrm{String}(3))=\mathbb{Z}/3\mathbb{Z}$... etc
A: The result is that a compact, connected simple Lie group $G$ has $\pi_3(G) = \mathbb{Z}$. Simple covering space or subgroups arguments should get you to $\mathrm{SO}(n)$ which is all that matters. For that matter start with the 1-connected $\mathrm{Spin}(n)$.
[OK, a short train ride later, now I'm home from work. To continue...]
The fibre of the 3-connected cover is a 2-type, and in the case of $\mathrm{Spin}(n)$ this is a $K(\mathbb{Z},2)$, so at the very least, $\mathrm{String}(n)$ can't be finite-dimensional. If one could construct a primitive[1] $PU(\mathcal{H})$-bundle on $\mathrm{Spin}(n)$ whose Dixmier-Douady classs was the generator $\langle -,[-,]\rangle \in H^3(\mathrm{Spin}(n),\mathbb{Z})$, then you would have an infinite-dimensional Lie group model for $\mathrm{String}(G)$ (here $\mathcal{H}$ is a infinite-dimensional separable Hilbert space, $PU(\mathcal{H})$ is then a smooth model for $K(\mathbb{Z},2)$, if we take the norm topology, making it a Banach Lie group).
([1] Primitive in the sense that for the group operations $G\times G\to G$ and $(-)^{-1}:G\to G$ there are bundle maps covering them.)
I don't know if this is possible or not, but I'm sure this idea has occurred to someone before, and since we haven't seen it, there might be a reason (well, I haven't seen it and everyone goes on about $\mathrm{String}_G$ only being a topological group).
