The key is that the matrix $U^{\text{new}}$ is again unitary, as you mention.
Let $A$ be a unitary matrix with $m$th row $(0,\ldots,0,a,0,\ldots,0)$.
Since $A$ is unitary, the product of $A$ with its transpose conjugate is the identity, that is, $A \overline{A^t} = I$. The $m$th diagonal element in $A \overline{A^t}$ is:
$$ 0 + \ldots + 0 + a \cdot \overline{a} + 0 +\ldots + 0 = a \cdot \overline{a} = |a|^2 $$
and this has to be one, so $|a|^2 = 1$.
Say the $m$th column is $(r_1,\ldots,r_k,a,r_{k+2},\ldots,r_n)$. Now $\overline{A^t}A=I$ must hold too and the $m$th diagonal element in $\overline{A^t}A$ is:
$$ |r_1|^2+ \ldots +|r_k|^2+|a|^2+|r_{k+2}|^2+\ldots+|r_n|^2$$
and this has to be one. Since $|a|^2=1$, this means
$$ |r_1|^2+ \ldots +|r_k|^2+|r_{k+2}|^2+\ldots+|r_n|^2 = 0 $$
from where all the $r_i$ are zero, that is, the off-diagonal elements of the $m$th column are all zero.
The answer: the Cauchy interlacing theorem which is presented here for hermitian matrices but, that, of course, can be "downgraded" to real symmetric matrices.
Best Answer
It is almost never unitary except in the trivial case that the original unitary matrix is (up to permutation) a block matrix. An easy way to see this is that the rows and columns of a unitary matrix must be orthonormal, so in particular must have norm $1$ as vectors, so the rows and columns of a principal submatrix will typically have norm less than $1$ because you're omitting entries of the corresponding vectors. They'll only have norm $1$ if all the omitted entries are $0$.