Not a linear subspace (it does not contain $\mathbf 0$, since $w^\top 0 = 0 \neq c$). It is an affine subspace of dimension $d-1$: it equals $x_0+V$ for $V=w^\perp$ (dimension $d-1$ by rank-nullity) and any particular solution $x_0$ with $w^\top x_0=c$.

Dimension $1$: $\text{span}\{r_f\} = \{\lambda r_f : \lambda \in \mathbb{R}\}$ is a 1-dimensional linear subspace, and a “direction” is simply this subspace together with a choice of unit-norm generator (the sign/normalization convention, not an intrinsic part of the subspace itself).

Since $\dim V = k$, its orthogonal complement $V^\perp$ has dimension $d-k$. Pick an orthonormal basis $w_1,\ldots,w_{d-k}$ of $V^\perp$. Then $V = \bigcap_{i=1}^{d-k} \{x : w_i^\top x = 0\}$: any $x\in V$ satisfies $w_i^\top x=0$ for all $i$ (since $w_i \perp V$), and conversely any $x$ orthogonal to all of $V^\perp$‘s basis vectors is orthogonal to all of $V^\perp$, hence lies in $(V^\perp)^\perp = V$. Each $\{x:w_i^\top x=0\}$ is a hyperplane through the origin, so $V$ is the intersection of $d-k$ such hyperplanes.

The unit sphere $S^{d-1} = \{r:\|r\|=1\}$ is not closed under addition or scalar multiplication (e.g. $r+r = 2r$ has norm $2\neq1$, so $r+r\notin S^{d-1}$ even though $r\in S^{d-1}$), so it fails to be a linear subspace. It is also not an affine subspace: affine subspaces are unbounded (whenever $\dim\geq1$) or single points, but $S^{d-1}$ is bounded and, for $d\geq2$, is not a single point — a curved manifold (Sphere), not a flat linear structure. Each individual $r$ spans a 1D linear subspace $\text{span}\{r\}$, but the collection of all such $r$ (the sphere itself) is a fundamentally different, curved object.

$\max(0,\, k_1+k_2-d) \;\leq\; \dim(V_1\cap V_2) \;\leq\; \min(k_1,k_2)$. Upper bound: $V_1\cap V_2 \subseteq V_1$ and $\subseteq V_2$, so its dimension is at most the smaller of the two. Lower bound: from the dimension formula $\dim(V_1+V_2) = \dim V_1+\dim V_2-\dim(V_1\cap V_2)$, and since $V_1+V_2\subseteq\mathbb{R}^d$ forces $\dim(V_1+V_2)\leq d$, rearranging gives $\dim(V_1\cap V_2) = k_1+k_2-\dim(V_1+V_2) \geq k_1+k_2-d$; combined with $\dim(V_1\cap V_2)\geq 0$ always, this gives the stated lower bound. Both bounds are achieved: e.g. two generic subspaces achieve the lower bound, and $V_1\subseteq V_2$ achieves the upper bound with equality.

Idempotence: $P_{V^\perp}$ is an orthogonal projection onto $V^\perp$; for any orthogonal projection $P$, $P(Px) = Px$ since $Px \in$ range$(P)$ already, and a projection fixes its own range pointwise (a defining property of orthogonal — indeed any idempotent — projections, $P^2=P$). Second claim: since $V'\subseteq V$, we have $V^\perp \subseteq (V')^\perp$ (larger subspace has smaller, more restrictive orthogonal complement… precisely: taking orthogonal complements reverses inclusions). $P_{V^\perp}x \in V^\perp \subseteq (V')^\perp$, and since $(V')^\perp$‘s projection fixes every vector already inside $(V')^\perp$ (same idempotence-type fact, applied to a vector already in the target subspace rather than to a repeated application of the same projection), $P_{(V')^\perp}(P_{V^\perp}x) = P_{V^\perp}x$.