It is
enough to show that $X=\mathbb A^2_k\setminus \lbrace 0\rbrace$ is not affine since $\mathbb A^2_k \:$ is affine.
First proof of non-affineness
The key point is that the restriction map $\Gamma(\mathbb A^2_k,\mathcal O_{\mathbb A^2_k})=k[T_1,T_2] \to \Gamma(X,\mathcal O_X )$ is bijective.
This is the analogue of the Hartogs phenomenon in several complex variables and in algebraic geometry results from the fact that for a normal ring $A$ we have $A=\cap_{\mathfrak p} A_{\mathfrak p}$, where the intersection is over primes of height $1$.
Now if $X$ were affine we would have the canonical isomorphism of schemes
$X\stackrel {\cong}{\to} Spec(\Gamma(X,\mathcal O_X ))=Spec (k[T_1,T_2])=\mathbb A^2_k$
which is false since the origin of $\mathbb A^2_k$ is not in $X$.
Edit: Second proof of non-affineness
Consider the open covering $\mathcal U$ of $X$ by the two open affine subsets $U_1=D(T_1)$ and $U_2=D(T_2)$, i.e. those open subsets determined by $T_1\neq0$ and $T_2\neq0$.
The covering is a Leray covering for $\mathcal O$ (acyclic opens with acyclic intersection), since $U_1, U_2$ and $U_1\cap U_2$ are affine.
Hence by Leray's theorem we have $H^1(X,\mathcal O)=\check {H}^1(\mathcal U,\mathcal O)$ and thus $H^1(X,\mathcal O)$ is the cohomology of the complex
$\Gamma(U_1,\mathcal O)\times \Gamma(U_2,\mathcal O)\to \Gamma(U_1\cap U_2,\mathcal O)\to 0$ where the non-trivial map is
$$k[T_1,T_1^{-1},T_2]\times k[T_1,T_2,T_2^{-1}] \to k[T_1,T_1^{-1},T_2,T_2^{-1}]:(f,g)\mapsto g-f$$
Thus $H^1(X,\mathcal O)=\oplus _{i,j\lt 0} \;\; k \cdot T^{i}T^{j}$, an infinite-dimensional $k$-vector space, in sharp contrast to Serre's theorem stating that positive dimensional cohomology groups of coherent sheaves on affine schemes are zero.
Other Edit: Third proof of non-affineness
If $k=\mathbb C $ let us show that $X$ is not affine by proving that its underlying holomorphic manifold $X_{hol}$ is not Stein.
Indeed suppose it were and consider the discrete closed subset $D=\lbrace (1/n,0): n=1,2,3,...\rbrace\subset X$.
Since $D$ is a $0$-dimensional submanifold the restriction map $\Gamma(X_{hol}, \mathcal O_{X_{hol}})\to \Gamma(D, \mathcal O_D)$ would be surjective.
On the other hand, by Hartogs's theorem the restriction map $\Gamma(\mathbb C^2,\mathcal O_{\mathbb C^2}) \to \Gamma(X_{hol},\mathcal O_{X_{hol}}) $ is also surjective so that by composition we would conclude if $X_{hol}$ were Stein to the surjectivity of the restriction map
$$ \Gamma(\mathbb C^2,\mathcal O_{\mathbb C^2}) \to \Gamma(D, \mathcal O_D): f\mapsto f_0=f\mid D $$
But this is clearly false because a holomorphic function $f_0:D\to \mathbb C$ is an arbitrary function, and if $f_0$ is not bounded [ for example $f_0(1/n,0)=n$ ] it cannot be the restriction of a holomorphic function $f:\mathbb C^2\to \mathbb C$.
One of the fundamental theorems in quantum mechanics is Wigner's theorem. It says
that a map preserving the inner product on a complex Hilbert space is unitary or
anti-unitary. It's a fairly easy consequence of the generalization of the fundamental theorem of projective geometry to infinite-dimensional spaces, which itself follows easily from the theorem as you stated it.
From the perspective of the mathematical foundations of quantum mechanics, it
shows that observables have to correspond to unitary or anti-unitary operators.
A map as above is called a symmetry and the citation for Wigner's 1963 Nobel prize in Physics included the phrase "for his contributions to the theory of the atomic nucleus and the elementary particles, particularly through the discovery and application of fundamental symmetry principles", see www.nobelprize.org/nobel_prizes/physics/laureates/1963/
Best Answer
The thing is, you might want to get some topology in the picture. In fact, if you do not, you can choose any bijection between the sphere and a $\mathbb R$-vector space, and you end up with a structure of vector space on your "sphere" (by transporting the structure). My point is, there exist such $\varphi$, but what you really want is not for $ \varphi_O$ to be only bijective : if your space already has a shape, you want it to be a homeomorphism.
And there is no homeomorphism between the sphere and a $\mathbb R$-vector space (for example because a vector space is contractible - you can shrink it continuously into a point - whereas the sphere is not ; you can look that up in any basic course of algebraic topology)