Basically the reason is that the classical conformal symmetry no longer holds at the quantum level due to the presence of the trace anomaly. More precisely, the tracelessness of the quantum stress-energy tensor is incompatible with the normal ordering needed to define it. By cohomological reasons, the trace of the stress-energy tensor, although non-vanishing, must be a central element of the "quantized Witt algebra", i.e. it must commute with all its generators. This implies that the "quantized Witt algebra" must be a non-trivial central extension of the Witt algebra, which in this particular case must be unique up to isomorphism (thanks to user106422 for recalling this point below) - that is, a Virasoro algebra.
An exposition I particularly like on such matters is the little book of Martin Schottenloher, "A Mathematical Introduction to Conformal Field Theory" (2nd. edition), Lecture Notes in Physics 759 (Springer--Verlag, 2008).
Edit (June 15th 2022): this comes more than six years late, but only now have I managed to get time to think about Peter Kravchuk's two (correct) points in his comments. I apologize to all.
- Indeed the trace anomaly only appears in curved space-time. More precisely, if your reference state has a short-distance behavior similar to that of the Minkowski vacuum state (in the case of Lagrangian free fields, this reduces to the Hadamard property for the two-point function in that state), the expectation value of the trace of the normal-ordered stress-energy tensor (w.r.t. the aforementioned reference state by e.g. point splitting and exploiting the operator product expansion) is state-independent and proportional to both the central charge and the scalar curvature (even if e.g. in the case of Lagrangian free fields you conformally couple the field to the curvature so as to keep the conformal symmetry of the model at the classical level). In other words, the trace anomaly is always there; we just do not see it where the scalar curvature vanishes. The role of the latter is to introduce a scale into the model through the coupling of the field dynamics with the background metric. Curvature, however, is not the only way to do this; another one is to impose boundary conditions on the conformal field theory - the Casimir energy is then also seen to be proportional to the central charge. As a rule, the central charge manifests itself physically as a way the conformal field model reacts to the introduction of an external macroscopic scale by means of a partial (anomalous, not spontaneous) breakdown of that symmetry. A more detailed discussion on this may be found e.g. in pp. 138-146 of the book by P. Di Francesco, P. Mathieu and D. Sénéchal, Conformal Field Theory (Springer-Verlag, 1997).
- In four dimensions the conformal Lie algebra is (semi)simple and therefore has no nontrivial central extensions, but both the trace anomaly in the presence of curvature and the Casimir energy in the presence of boundary conditions persist, so what changes? Well, recall that in four dimensions conformal transformations (in Lorentz signature) are composites of Poincaré transformations, (rigid) scale transformations and the so-called special conformal transformations, whereas the group of conformal transformations in two dimensions is much larger than that (in fact, infinite dimensional!). In Euclidean signature, (one of the two chiral components of) the conformal subgroup generated in the same way as in higher dimensions consists of Möbius transformations on the complex plane and therefore equals the semisimple Lie group $SL(2,\mathbb{R})$. The corresponding "rigid" (Möbius) part of the Witt Lie algebra (generated by $l_0$, $l_1$ and $l_{-1}$), which generates all Möbius transformations, sees no change in the commutation relations among their generators when we perform the central extension to obtain the Virasoro algebra, since the central part of $[l_n,l_m]$ is then given by $\frac{c}{12}(n^3-n)\delta_{m+n,0}$, where $c$ is the central charge. In other words, the Möbius Lie subalgebra does not "see" $c$, just like in higher dimensions. Nonetheless, the appearance of the scalar curvature in the trace anomaly formula shows that this anomaly really only affects "local" (i.e. non-"rigid") conformal transformations. In the case of boundary conditions, they affect the field algebra itself (e.g. in the case of free fields the causal commutator Green function must respect the boundary conditions) and hence also the normal ordering of the stress-energy tensor through shifting by central elements of the field algebra. Needless to say, usually global (resp. "rigid" = Möbius) conformal transformations (resp. in two dimensions) do not survive either the introduction of curvature or boundary conditions, so the role of the ("rigid") conformal Lie (sub)algebra in the present matter becomes moot since they no longer generate symmetries of the field theory anyway.
Schottenloher's chapter 3 and 4 are somewhat "non-linear" in that a lot of chapter 3 only makes sense after you've seen what chapter 4 does with it. In this particular case, sure, theorem 3.10 as written would apply to $G=\mathrm{SO}(3)$, and you would get some $E$ as a central extension. But this is a useless observation, because we don't have any machinery that would tell us something about this $E$! What you really want is that $E$ splits as $E\cong G\times \mathrm{U}(1)$ so that the lifting $S:E\to\mathrm{U}(\mathbb{H})$ reduces to a proper linear representation $S_2 : G\to\mathrm{U}(\mathbb{H})$.
Chapter 4 presents Bargmann's theorem as the main tool for this - the sequence splits for a $G$ that is simply-connected and has $H^2(\mathfrak{g},\mathbb{R}) = 0$. As the universal cover of $\mathrm{SO}(3)$, $\mathrm{SU}(2)$ is simply-connected and so Bargmann's theorem will apply to it, which is why Schottenloher is using it in this example.
You never need to worry about applying theorem 3.10 to groups that aren't simply connected: Every projective representation of a group lifts to a projective representation of its universal cover and in fact all projective representations of the cover are also projective representations of $G$ because the universal cover is just a central extension by the fundamental group $\pi_1(G)$, so we can always apply this theorem to the universal cover to have a chance of applying Bargmann's theorem afterwards.
If "the universal cover is just a central extension by the fundamental group" doesn't convince you because we're actually talking about central extensions by $\mathrm{U}(1)$, not discrete fundamental groups, the rest of this answer is for you.
Explicitly, let $\tilde{G}$ be the universal cover, then we have the projection
$$ \hat{\pi} : \mathrm{U}(1)\times \tilde{G} \to G$$
and when $G$ is compact and not the product of smaller Lie groups, $\pi_1(G)\cong \mathbb{Z}_m$ for some prime $m$. There is an embedding $\iota : \mathbb{Z}_m\to \mathrm{U}(1)$ mapping the elements of $\mathbb{Z}_m$ to the $m$-th roots of unity, and so we have a "diagonal" inclusion
$$ \mathbb{Z}_m\to \mathrm{U}(1)\times \tilde{G}, \gamma\mapsto (\iota(\gamma), \gamma)$$
Let's call the image of this inclusion $A$. Then clearly $A$ lies in the kernel of the projection
$$ \hat{\pi} :\mathrm{U}(1)\times \tilde{G}\to G, (x,g) \mapsto \pi(g)$$
where $\pi$ is the normal projection from the universal cover, and so we can turn the central extension
$$ 1\to \mathrm{U}(1)\times \pi_1(G) \to \mathrm{U}(1)\times \tilde{G} \overset{\hat{\pi}}{\to} G \to 1$$
into
$$ 1\to (\mathrm{U}(1)\times \pi_1(G))/A \cong \mathrm{U}(1) \to (\mathrm{U}(1)\times \tilde{G})/A \to G\to 1$$
where the thing in the middle is a "non-trivial" central extension of $G$ but only in the same way that $\tilde{G}$ is a non-trivial central extension of $G$, i.e. this doesn't have anything to do with the Lie algebra or its $H^2(\mathfrak{g},\mathbb{R})$ - on the level of the algebras, this extension is trivial.
Best Answer
To clarify - $\mathbb P\mathcal H$ is not a vector space, and so I'm not so sure what a unitary operator on $\mathbb P\mathcal H$ would be. My assumption is that you are referring to the ray product $(\Phi,\Psi)$ between elements of $\mathbb P\mathcal H$, defined via $$(\Phi,\Psi):= \frac{|\langle\phi,\psi\rangle|}{\Vert\phi\Vert \ \Vert\psi\Vert}$$ where $\phi,\psi\in \mathcal H$ are any two representatives of the equivalence classes $\Phi$ and $\Psi$. A symmetry transformation is then a map $T:\mathbb P\mathcal H \rightarrow \mathbb P\mathcal H$ such that $(T\Phi,T\Psi)=(\Phi,\Psi)$. This looks rather like the definition of a unitary operator on a Hilbert space, but there is no notion of linearity present here.
Wigner's theorem tells us that any such $T$ can be induced by some unitary or antiunitary operator on $\mathcal H$, which descends to a transformation on $\mathbb P\mathcal H$ in the obvious way. By extension, if we want to represent the action of a symmetry group $G$ rather than just a single transformation, we might seek linear representations $\rho:G\rightarrow U(\mathcal H)$. However, requiring $\rho$ to be a true group homomorphism is generally too strong; if it is a homomorphism up to a phase - i.e. $\rho(g)\rho(h)=\rho(gh)C(g,h)$ for some phase factor $C(g,h)$ - then it also induces a well-defined group action on $\mathbb P\mathcal H$ because the extra phase factor is lost during the projectivization from $\mathcal H$ to $\mathbb P\mathcal H$.
The takeaway is that the action of a symmetry group $G$ on $\mathbb P\mathcal H$ can be understood through a projective unitary representation of $G$ on $\mathcal H$, so we need to understand the latter if we want to understand how to implement symmetry groups in our theories.
If $\rho:G\rightarrow \mathrm{Aut}(\mathbb P\mathcal H)$ is a group action whose elements are symmetry transformations as defined above, then Wigner's theorem tells us that that it lifts to a projective unitary representation $\hat \rho:G\rightarrow \mathbb PU(\mathcal H)$. It may lift to a genuine representation - i.e. one might be able to arrange that the phase factors $C(g,h)$ referenced above are all equal to $1$ - but in general this cannot be done.
As it turns out, projective representations of $G$ can be put into one-to-one correspondence with (genuine) linear representations of its central extensions (the same is true for the Lie algebra $\mathfrak g$). Since linear representations are far nicer to work with, this is what we tend to study.
ACuriousMind has written a wonderful answer on this subject which you can find here. I doubt any summary I could provide would do it justice, so I suggest reading it directly if you haven't already done so.