Let $M \in \mathbb{R}^{m \times n}$ be a matrix of rank $r$ with compact SVD $M=U\Sigma V^T$ ($U \in \mathbb{R}^{m \times r}$ and $V \in \mathbb{R}^{n \times r}$ are semi-unitary matrices and $\Sigma$ only contains the strictly positive spectral values of $M$). Denote by $U' \in \mathbb{R}^{m \times r}$ and $V' \in \mathbb{R}^{n \times r}$ two other arbitrary semi-unitary matrices and let $U'_{\perp} \in \mathbb{R}^{m \times (m-r)}$ (resp. $V'_{\perp} \in \mathbb{R}^{n \times (n-r)}$) be a matrix made of orthonormal vectors completing $U'$ (resp $V'$) in order to obtain an orthonormal basis of $\mathbb{R}^{m}$ (resp. $\mathbb{R}^{n}$).
It is claimed in a paper that I am reading that $$\left\|U'_{\perp}\left(U^{'T}_{\perp}MV'_{\perp}\right)V^{'T}_{\perp}\right\|_{\infty} \leq \left\|U'U'^T-UU^T\right\|_{2 \to \infty} \left\|M\right\|_{op} \left\|V'V'^T-VV^T\right\|_{2 \to \infty},$$ where the infinity norm is the maximal entry of a matrix, the 'op' norm is the operator norm, and the two-to-infinity norm is largest euclidean norm of the rows of a matrix.
Using the fact that $\left\|ABC\right\|_{\infty} \leq \left\|A\right\|_{2 \to \infty}\left\|B\right\|_{op}\left\|C^T\right\|_{2 \to \infty}$ for any $A,B,C$ (this follows from Proposition 6.5. of this article), I am able to show that $$\left\|U'_{\perp}\left(U^{'T}_{\perp}MV'_{\perp}\right)V^{'T}_{\perp}\right\|_{\infty} \leq \left\|U'_{\perp}U^{'T}_{\perp}U\right\|_{2 \to \infty} \left\|M\right\|_{op} \left\|V'_{\perp}V^{'T}_{\perp}V\right\|_{2 \to \infty},$$ but I fail to see how to obtain the claimed bound from this result.
Observe that since $U_{\perp}'U_{\perp}^{'T}$ is the matrix such that $[U', U_{\perp}']$ is orthogonal, it holds that $U_{\perp}' U_{\perp}{'T} = (I - U'U'^{T})$, and similarly for $V'$. Therefore, \begin{align*} \| U_{\perp}' U_{\perp}'^t U \|_{2\to\infty} \| M \|_{op} \| V_{\perp}' V_{\perp}'^T V \|_{2,\infty} &= \| (I - U'U'^T) U \|_{2\to\infty} \| M \|_{op} \|(I - V' V'^T) V \|_{2,\infty} \\ &= \| U - U'U'^T U \|_{2\to\infty} \| M \|_{op} \|V - V' V'^T V \|_{2,\infty}. \end{align*} Furthermore, $\|\cdot\|_{2,\infty}$ is invariant to right multiplication by a matrix with orthogonal rows. Hence $\|U - U' U'^T U \|_{2,\infty} = \| UU^T - U' U'^T U U^T \|_{2,\infty}$, and similarly for the term with $V$ (I believe this is in the paper you cited). In addition, $UU^T = UU^T UU^T$ since it is an orthogonal projection. Hence, \begin{align*} \| U - &U'U'^T U \|_{2\to\infty} \| M \|_{op} \|V - V' V'^T V \|_{2,\infty}\\ &= \| UU^T - U' U'^T U U^T\|_{2,\infty} \| M \|_{op} \| VV^T - V' V'^T V V^T \|_{2,\infty} \\ &= \| UU^T UU^T - U' U'^T U U^T \|_{2,\infty} \| M \|_{op} \| VV^T - V' V'^T V V^T \|_{2,\infty} \\ &\leq \| UU^T - U' U'^T \|_{2,\infty} \| M \|_{op} \| VV^T - V' V'^T \|_{2,\infty} \end{align*} where we have used the identity $\| AB \|_{2,\infty} \leq \| A \|_{2,\infty} \| B\|_{op}$ together with the fact that $\|UU^T \|_{op} \| VV^T \|_{op} = 1$.