Let $Y$ and $X$ be positive semidefinite matrices of size $d \times d$. We define $\sqrt{Y}$ and $\sqrt{X}$ as the unique symmetric positive semidefinite matrices that satisfy $(\sqrt{Y})^2 = Y$ and $(\sqrt{X})^2 = X$. I need to show that there exists a constant $C>0$ (that may depend on $d$) such that $$|| \sqrt{X}-\sqrt{Y}||^2 \leq C ||X-Y||. $$ The norm is defined as $||A|| = \sqrt{Trace(AA^*)}.$
Here's what I first did: By considering the eigendecomposition $Y = H \Lambda H^*$, we have $\sqrt{Y} = H \sqrt{\Lambda} H^*$ and $$|| \sqrt{X}-\sqrt{Y}||^2 = Trace(( H \sqrt{\Lambda} H^* - \sqrt{X})(H \sqrt{\Lambda} H^* - \sqrt{X})^* ) = Trace(H ( \sqrt{\Lambda} - H^*\sqrt{X} H )(\sqrt{\Lambda} - H^*\sqrt{X} H)^* H^*) = Trace(( \sqrt{\Lambda} - H^*\sqrt{X} H )(\sqrt{\Lambda} - H^*\sqrt{X} H)^*) $$ since $H H^* = I_d.$ In other words, we can assume that $Y$ is diagonal, without loss of generality. Note that $H^*\sqrt{X} H$ is also positive semidefinite.
Then: $$ Trace(( \sqrt{Y} - \sqrt{X} )(\sqrt{Y} - \sqrt{X})^*) = Trace( Y - X + 2 X -\sqrt{X} \sqrt{Y} - \sqrt{Y} \sqrt{X} )$$ $$ Trace( Y - X ) + 2 Trace( X -\sqrt{X} \sqrt{Y} ) = Trace( Y - X ) + 2 Trace( \sqrt{X} ( \sqrt{X} - \sqrt{Y}) ) $$ $$ \leq ||Y-X|| + 2 Trace( \sqrt{X} ( \sqrt{X} - \sqrt{Y}) ) $$
Then what?