I was trying to find the least value of |sin⁡x|+|cos⁡x| and applied the above inequality...

It doesn't hold iff x and y have the same sign. As for your problem remember that if ${\displaystyle a\le 1}$ then ${\displaystyle a^2\le a}$ so
${\displaystyle \left|\sin x\right|+\left|\cos x\right|\ge {\left|\sin x\right|}^2+{\left|\cos x\right|}^2=1}$
and for upper bound (remember ${\displaystyle {(a+b)}^2\le 2(a^2+b^2)}$):
${\displaystyle \left|\sin x\right|+\left|\cos \right|x\le \sqrt{2({\left|\sin x\right|}^2+{\left|\cos x\right|}^2)}=\sqrt{2}}$

${\displaystyle |\sin x+\cos x|\le \left|\sin x\right|+\left|\cos x\right|}$ is correct.
But, for example, if x=0 then both sides are 1; thus we cannot conclude ${\displaystyle \sqrt{2}\le \left|\sin x\right|+\left|\cos x\right|}$

Let ${\displaystyle \{\begin{array}{c}f\left(x\right)=\sin x+\cos x\\ g\left(x\right)=|\sin \left(x\right)|+|\cos \left(x\right)|\end{array}}$
Notice that $g(x+\frac{\pi }{2})=|\cos (x)|+|\sin (x)|=g(x)$ so
g is periodic of period $\frac{\pi }{2}$ and we can study it on $[0,\frac{\pi }{2}]$.
But on this interval both sin and cos are positive, so we can get rid of absolute values and g(x)=f(x) on $[0,\frac{\pi }{2}].$
You now can use the addition formulas to transform f to $f(x)=\sqrt{2}\sin (x+\frac{\pi }{4})$ and see that $f(x)$ has values between 0 and $\sqrt{2}$ while g(x) has values between 1 and $\sqrt{2}$.