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The orthogonal complement $x^{\perp }$ is infinite-dimensional: otherwise $X=\mathrm{span}\{x\}\oplus x^{\perp }$ would be finite-dimensional. Inductively choose an orthonormal sequence $(e_n)$ in $x^{\perp }$, each new vector being orthogonal to the preceding finite linear span. The Bessel inequality gives ${\sum }_n|⟨e_n,z⟩{|}^2\le \Vert z{\Vert }^2$ for every $z\in X$, so $⟨e_n,z⟩\to 0$. Thus $(e_n)$ is a weakly null orthonormal sequence.

SetBy orthogonality, $\Vert x_n{\Vert }^2=\Vert x{\Vert }^2+(1-\Vert x{\Vert }^2)=1$, while $⟨x_n-x,z⟩\to 0$ for every $z$. Hence $x_n\rightharpoonup x$. When $\Vert x\Vert =1$, this formula simply gives the constant sequence $x_n=x$. In particular, the weak closure of the unit sphere contains the whole closed unit ball.

No. Take the Hilbert space $H={\ell }^2$ and its standard orthonormal sequence $(e_n)$. Letthe unit sphere. This set is norm closed but is not convex. By the Bessel inequality, for every $y=(y_j)\in {\ell }^2$,so $e_n\rightharpoonup 0$; equivalently, this is a weakly null orthonormal sequence. Every $e_n$ lies in $C$, but the weak limit $0$ does not. Hence convexity is essential.

No. In $H={\ell }^2$, let $x_n=e_n$ be the standard orthonormal sequence. For any $y=(y_j)\in {\ell }^2$,so $e_n\rightharpoonup 0$. However,for $n\ne m$. No subsequence is Cauchy, hence none converges strongly. This is a weakly null orthonormal sequence.