Showing that matrix $Q=UV^T$ is the nearest orthogonal matrix to $A$.

Note that, as stated, the question only makes sense if $n=m$, because in the singular value decomposition of $A$, $U$ will be $m\times m$ and $V^T$ will be $n\times n$.

Because the Frobenius norm is unitarily invariant, you have $$ \|A-Q\|_F=\|U\Sigma V^T-Q\|_F=\|\Sigma-U^TQV\|_F. $$ But the orthogonal (or the unitary) matrices form a group, so you want to minimize $$ \|\Sigma-Q\|_F $$ over all orthogonal matrices. You have \begin{align} \|\Sigma-Q\|_F^2&=\sum_k(\Sigma_{kk}-Q_{kk})^2+\sum_{j\ne k}Q_{kj}^2\\ \ \\ &=\sum_k(\Sigma_{kk}^2+Q_{kk}^2-2\Sigma_{kk}Q_{kk})+\sum_{j\ne k}Q_{kj}^2\\ \ \\ &=\sum_k(\Sigma_{kk}^2-2\Sigma_{kk}Q_{kk})+\sum_{j,k}Q_{kj}^2\\ \ \\ &=\text{Tr}(\Sigma^2)+\text{Tr}(Q^TQ)-2\sum_k\Sigma_{kk}Q_{kk}\\ \ \\ &=\text{Tr}(\Sigma^2)+n-2\sum_k\Sigma_{kk}Q_{kk} \end{align} To minimize this quantity over $Q$, since the entries of $\Sigma$ are non-negative and $Q_{kk}\in[-1,1]$, we need to choose $Q_{kk}=1$ for all $k$, which makes $Q=I$.

So the minimum is $$ \|\Sigma-I\|_F=\|U\Sigma V^T-UV^T\|_F=\|A-UV^T\|_F. $$