We establish the following manifestation of the Cauchy-Schwartz inequality.
\begin{align}
\text{tr}(CD)&=\sum_{ij}C_{ij}D_{ji} \\
&\le\Big(\sum_{ij}C_{ij}^2\Big)^\frac12\Big(\sum_{ij}D_{ij}^2\Big)^\frac12 \\
&=\big(\text{tr}(C^TC)\big)^\frac12\big(\text{tr}(D^TD)\big)^\frac12=\|C\|\|D\|,
\end{align}
for any real square matrices $C$ and $D$.


Perform the singular value decomposition of $X=ASB^T$ where $S$ is the diagonal matrix of the singular values of $X$ and $A$ and $B$ are the associated orthogonal matrices. Apply the above proposition, we have
$$
\|X\|_\sigma = \text{tr}(S) = \text{tr}(A^TUV^TB)\le \|A^TU\|\|V^TB\|=\|U\|\|V\|\le \frac12(\|U\|^2+\|V\|^2).
$$

The equalities are achieved when $U=A\sqrt S$ and $V=B\sqrt S$.

We obtain the desired result.