We use Von Neumann's trace inequality to prove the proposition.
\begin{align}
\|UV^T\|_\sigma&=\text{tr}\sqrt{VU^TUV^T}=\text{tr}\big(\sqrt{U^TU}\sqrt{V^TV}\big) \\
&\le \sum_{i=1}^n\sigma_i\big(\sqrt{U^TU}\big)\sigma_i\big(\sqrt{V^TV}\big) \quad \text{(Von Neumann's trace inequality)}\\
&=\sum_{i=1}^n\sigma_i(U)\,\sigma_i(V) \\
&\le\Big(\sum_{i=1}^n \sigma_i(U)^2\Big)^\frac12
\Big(\sum_{i=1}^n \sigma_i(V)^2\Big)^\frac12 \quad\text{(Cauchy-Schwartz inequality)} \\
&= \|U\|\|V\| \le \frac12\big(\|U\|^2+\|V\|^2\big), \quad\text{(arithmetic-geometric mean inequality)}
\end{align}
where $\sigma_1(M)\ge\sigma_2(M)\ge\cdots\sigma_n(M)$ are the singular values of the matrix $M$.
In fact, we can dispense with Von Neumann's inequality and simplify the above inequality with a direct application ofestablish the following menifastationmanifestation 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 $D$.
Apply the above inequality to $C=\sqrt{U^TU}$ and $D=\sqrt{V^TV}$, we have
$$\text{tr}\big(\sqrt{U^TU}\sqrt{V^TV}\big)\le \|U\|\|V\|.$$
Now takePerform 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. For $U=A\sqrt S$ and $V=B\sqrt S$,
$$\|U\| = \sqrt{\text{tr}(ASA^T)}=\sqrt{\text{tr}\,S}$$
andApply the same for $V$. We alsoabove proposition, we have
$$\|UV\|_\sigma=\text{tr}\,S$$
So both $\|U\|\|V\|$ and $\frac12\big(\|U\|^2+\|V\|^2\big)$ achieve its minimum at$$
\|X\|_\sigma = \text{tr}(S) = \text{tr}(A^TUV^TB)\le \|A^TU\|\|V^TB\|=\|U\|\|V\|\le \frac12(\|U\|^2+\|V\|^2).
$$
The equalities are achieved when $U=A\sqrt S$ and $V=B\sqrt S$.
We obtain the desired result.
