Since $H = I- 2ww^*$ is hemitian, the singular values of $H$are just the absolute values of the eigenvalues of $H$.
Notice that if $u\perp w$, then $$Hu = u - 2ww^*u = u-2\langle u,w\rangle w = u$$
so $u$ is an eigenvector for $H$ with eigenvalue $1$. If you are working in $\mathbb{C}^n$, you can pick $n-1$ linearly independent vectors orthogonal to $v$ so the multiplicity of the eigenvalue $1$ is at least $n-1$.
On the other hand, we have $$Hw = w - 2ww^*w = w - 2\|w\|^2w = w-2w = -w$$
so $w$ is the eigenvector for $H$ with eigenvalue $-1$. Therefore, the only eigenvalues are $\pm 1$ so the desired singular value is $1$.