$\newcommand{\R}{\mathbb{R}}
\newcommand{\U}{\mathbf{U}}
\newcommand{\W}{\mathbf{W}}
\newcommand{\A}{\mathbf{A}}
\newcommand{\tr}{\mathrm{tr}}
\newcommand{\rank}{\mathrm{rank}}
$
Choosing an arbitrary basis, we may assume $V=\R^n$.
We will show that the mapping
\begin{align}
\mathbb{S}\big(\bigwedge\nolimits^k_s(\R^n)\big)&\to Gr_k(\R^n)\\
\sigma=u_1\wedge\ldots\wedge u_k &\mapsto\operatorname{span}\{u_1,\ldots,u_k\}=\{v\in\R^n\colon v\wedge\sigma=0\}
\end{align}
is continuous w.r.t. the angle metric on the Grassmannian and induced inner product metric on the exterior power. The following lemma implies this assertion. Here $\mathbb{S}(X)$ denotes the set of elements in $X$ of unit length and $\bigwedge\nolimits^k_s(\R^n)$ is the set of simple $k$-vectors.
Let $U,W\in Gr_k(\R^n)$ and choose orthonormal bases $u_1,\ldots,u_k$ and $w_1,\ldots,w_k$ of $U$ and $W$, respectively.
Let further $\U,\W\in\R^{n\times k}$ be the matrices with columns $(u_i)$ and $(w_i)$, respectively.
Finally, define the unit $k$-vectors $\sigma=u_1\wedge\cdots\wedge u_k$ and $\tau=w_1\wedge\cdots\wedge w_k$.
Lemma
\begin{align}
d(U,W) \leq \sqrt{k}\left\|\sigma-\tau\right\|_{\bigwedge^k(\R^n)}
\end{align}
Proof
First, note that by the Cauchy-Schwartz inequality
\begin{align}
\det(\U^T\W)=\langle\sigma,\tau\rangle_{\bigwedge^k(\R^n)}
\leq \|\sigma\|_{\bigwedge^k(\R^n)}\|\tau\|_{\bigwedge^k(\R^n)}=1. \tag{1}\label{eq:CS_ineq}
\end{align}
Secondly, the Euclidean operator norm of a linear operator $A$ is bounded by its Frobenius norm, i.e.
$\|A\| \leq \|A\|_F$. This follows from the fact that the Euclidean norm and the Frobenius norm of a vector coincide and the submultiplicativity of the Frobenius norm.
Therefore,
\begin{align}
d(U,W)^2
&= \left\| P_U-P_W \right\|^2
\leq \left\| P_U-P_W \right\|_F^2
=\left\| P_U\right\|_F^2 +\left\|P_W \right\|_F^2 -2\left\langle P_U,P_W \right\rangle_F.
\end{align}
Using that $P_U=\U^T\U$ and $P_W=\W^T\W$ are idempotent, we get that $\left\| P_U\right\|_F^2=\rank(P_U)=k$ and also $\left\| P_W\right\|_F^2=k$.
Thus,
\begin{align}
\left\| P_U-P_W \right\|_F^2
&=2k-2\tr(\U\U^T\W\W^T)
=k\big(2-\tfrac2k \tr((\U^T\W)^T\U^T\W)\big).
\end{align}
Due to the inequality of the arithmetic and geometric mean, it holds that
\begin{align}
\sqrt[k]{\det(\A)} \leq \tfrac1k \tr(\A)
\end{align}
for any matrix $\A\in\R^{k\times k}$ (because the trace is the sum and the determinant the product of the eigenvalues of $\A$).
If $\det(\A)\leq1$ then for any $k\geq2$
\begin{align}
\sqrt[2]{\det(\A)} \leq \tfrac1k \tr(\A). \tag{2}\label{eq:AGM}
\end{align}
Therefore we can estimate using \eqref{eq:CS_ineq} and \eqref{eq:AGM} for $\A=(\U^T\W)^T\U^T\W$
\begin{align}
\left\| P_U-P_W \right\|_F^2
&\leq k\big(2-2\sqrt{\det((\U^T\W)^T\U^T\W)}\big)
=k\big(2-2\det(\U^T\W)\big).
\end{align}
Note that we can always arrange for $\det(\U^T\W)$ to be positive be interchanging two columns of, say, $\U$. This is possible because we investigate the unoriented Grassmannian.
Finally, direct computation shows that
\begin{align}
\|\sigma-\tau\|_{\bigwedge^k(\R^n)}^2=\|\sigma\|_{\bigwedge^k(\R^n)}^2+\|\tau\|_{\bigwedge^k(\R^n)}^2 - 2 \langle\sigma,\tau\rangle_{\bigwedge^k(\R^n)}^2 = 2-2\det(\U^T\W)
\end{align}
which concludes the proof.
The next statement is not really needed but just included here for completion.
Corollary
For arbitrary $\sigma,\tau$ arbitrary $k$-vectors such that they span the subspaces $U$ and $W$, it holds that
\begin{align}
d(U,W)
\leq \sqrt{k}\left\|\tfrac{\sigma}{\left\|\sigma\right\|_{\bigwedge^k(\R^n)}}- \tfrac{\tau}{\left\|\tau\right\|_{\bigwedge^k(\R^n)}}\right\|_{\bigwedge^k(\R^n)}
\leq \sqrt{k}\frac{\left\|\sigma-\tau\right\|_{\bigwedge^k(\R^n)}}{\left\|\sigma\right\|_{\bigwedge^k(\R^n)}\left\|\tau\right\|_{\bigwedge^k(\R^n)}}
\end{align}
Proof
If $\sigma=a_1\wedge\cdots\wedge a_k$ for an arbitrary basis $(a_1,\ldots,a_k)$ of $U$ then we can change basis to an orthonormal basis $(u_1,\ldots,u_k)$, e.g. via QR-decomposition of $\A=\U\mathbf{R}$. Then $\sigma=\det(\mathbf{R})u_1\wedge\cdots\wedge u_k$ and $\left\|\sigma\right\|_{\bigwedge^k(\R^n)}=\det(\mathbf{R})$.
Then use the lemma for $\sigma/\left\|\sigma\right\|_{\bigwedge^k(\R^n)}$ and $\tau/\left\|\tau\right\|_{\bigwedge^k(\R^n)}$.
The second inequality follows from the fact that $2ab\leq a^2+b^2$ for any real numbers $a,b$.
