Bounding the minimum singular value of a block triangular matrix

Yes, this works. As I suggested in my comment above, we can rephrase the claim as $\|M\|^2 \le \|A\|^2 + \|B\|^2 + \|D\|^2$, and this we can check by direct calculation: Apply $M$ to $v=(x,y)^t$. Then $$ \|Mv\|^2 =\|Ax+By\|^2 + \|Dy\|^2 $$ and $$ \|Ax+By\|^2 \le (\|A\| \|x\|+\|B\|\|y\|)^2 \le (\|A\|^2 + \|B\|^2)(\|x\|^2+\|y\|^2) , $$ (by the Cauchy-Schwarz inequality on $\mathbb R^2$ in the last step) as desired.

It suffices to state the bound for $M$, as also noted in C. Remling's comment.

In particular, recall that $\|M\|^2 = \|M^*M\|$, whereby we have the bound \begin{equation*} \|M\|^2 = \left\|\begin{bmatrix} A^*A + B^*B & B^*D\\ D^*B & D^*D \end{bmatrix}\right\| \le \|A^*A + B^*B + D^*D\|, \end{equation*} where the final inequality holds as the block matrix involved is PSD.

One can even prove related bounds if norms other than the operator norm are desired. See for instance this interesting paper by K. Audenaert for bounds involving Schatten-$p$ norms (though it leads to somewhat different inequalities); while Bourin's paper provides such inequalities for all symmetric (i.e., unitarily invariant) norms.