Show that $|z - w| \geq \big||z|-|w|\big|$ given $z = x + iy$ and $w = u + iv$.
So far I have,
$$|z - w| \geq \big||z|-|w|\big|$$ $$\sqrt{(x - u)^2 + (y - v)^2} \geq |\sqrt{x^2 + y^2} -\sqrt{u^2 + v^2}|$$
At this point I get a little 'undone'.
I tried a few different routes, but none of them have come up with the desired result.
\begin{align*} \mid z-w \mid&=\sqrt{(z-w)\overline{(z-w)}}\\ &=\sqrt{(z-w)(\overline{z}-\overline{w})}\\ &=\sqrt{\mid z\mid^{2}+\mid w\mid^{2}-2\Re(z\overline{w})}\\ &=\sqrt{(\mid z\mid-\mid w\mid)^{2}+2\mid z\mid \mid w\mid-2\Re(z\overline{w})}\\ &=\sqrt{(\mid z\mid-\mid w\mid)^{2}+2\mid z\overline{w}\mid-2\Re(z\overline{w})}\\ &\geq\sqrt{(\mid z\mid-\mid w\mid)^{2}}\text{ \{since $\mid z\overline{w}\mid \geq \Re(z\overline{w})$ \}}\\ &=\bigg| \mid z\mid -\mid w\mid \bigg| \end{align*}
Without using the triangle inequality, we can write
$$\begin{align} |z+w|&=\sqrt{(z+w)\overline{(z+w)}}\\\\ &=\sqrt{|z|^2+|w|^2+2\text{Re}(z\bar w)}\\\\ &=\sqrt{(|z|-|w|)^2+2\overbrace{(|z|\,|w|+2\text{Re}(z\bar w))}^{\ge 0\,\,\text{since}\,\,|\text{Re}(z)|\le |z|}}\\\\ &\ge \sqrt{(|z|-|w|)^2}\\\\ &=||z|-|w|| \end{align}$$
as was to be shown!
Using the Triangle Inequality, we get $$|z|=|(z-w)+w|\leq |z-w|+|w|\tag 1$$ and $$|w|=|(w-z)+z|\leq |w-z|+|z|=|z-w|+|z|.\tag 2$$ Using $(1)$ we get $$|z|-|w|\leq |z-w|\tag 3$$ and using $(2)$ we get $$-(|z|-|w|)\leq |z-w|,$$ that is we get $$-|z-w|\leq |z|-|w|.\tag 4$$ Combining $(3)$ and $(4)$, we get $$-|z-w|\leq |z|-|w|\leq |z-w|.$$ Because $|z-w|\geq 0$, we get $$\big||z|-|w|\big|\leq |z-w|.$$
You probably know that $\vert a+b\vert\le\vert a\vert+\vert b\vert$ for all $(a,b)\in\mathbb{C}^2$.
Now, given $(z,w)\in\mathbb{C}^2$ :
$$\vert z\vert=\vert(z-w)+w\vert\le\vert z-w\vert+\vert w\vert$$
Hence :
$$\vert z\vert-\vert w\vert\le\vert z-w\vert$$
And by symmetry, we also have :
$$\vert w\vert-\vert z\vert\le\vert w-z\vert$$
which can be written :
$$-\left(\vert z\vert-\vert w\vert\right)\le\vert z-w\vert$$
Finally :
$$\left|\vert z\vert-\vert w\vert\right|\le\vert z-w\vert$$as desired.
Other answers address your issue in a better way but I think it will be nice if a proof by your line of thought is given.
$$\sqrt{(x - u)^2 + (y - v)^2} \geq |\sqrt{x^2 + y^2} -\sqrt{u^2 + v^2}|$$
Since both sides are $>0$,
$$(x - u)^2 + (y - v)^2 \geq x^2 + y^2 - u^2 + v^2 -2\sqrt{x^2 + y^2}\sqrt{u^2 + v^2}$$
$$(x^2 + u^2 -2xu) + (y^2 + v^2 - 2yv) \geq x^2 + y^2 + u^2 + v^2 -2\sqrt{x^2 + y^2}\sqrt{u^2 + v^2}$$
$$-xu - yv \geq -\sqrt{x^2 + y^2}\sqrt{u^2 + v^2}$$
$$\sqrt{x^2 + y^2}\sqrt{u^2 + v^2} -xu - yv \geq 0$$
$$\sqrt{x^2u^2 + x^2v^2 + y^2u^2 + y^2v^2} -xu - yv \geq 0$$
$$x^2u^2 + x^2v^2 + y^2u^2 + y^2v^2 \geq x^2u^2 + y^2v^2 + 2xyuv $$
$$x^2v^2 + y^2u^2 \geq 2xyuv \tag{*}$$
If you remember (*) is golden AM-GM inequality.