Prove by contradiction, that for every $x ∈ [ π /2, π ]$, $\sin x − \cos x \ge 1$ I feel like I made some mistake along the way. Can somebody check, and explain if it's correct. The book provided different solution to this one.
For the sake of contradiction suppose that there exists $x\in[\frac{\pi}{2}, \pi]$, $\sin (x)-\cos(x)<1$. With algebraic and trig manipulations we get:
$$\sin(x)<1+\cos(x)$$
$$\sin^2(x)<1+2\cos(x)+\cos^2(x)$$
$$1-\cos^2(x)<1+2\cos(x)+\cos^2(x)$$
$$-2\cos^2(x)<2\cos(x)$$
$$-2\cos^2(x)-2\cos(x)<0$$
Dividing by $-2\cos(x)$ leads to:
$$\cos(x)+1<0$$
Since we know that $\cos(x)$ on the interval $[\frac{\pi}{2}, \pi]$ is between $0$ and $1$, $\cos(x)+1$ can't be lower than $0$. Thus contradiction.
 A: 
For the sake of contradiction suppose that there exists $x\in\left[\frac{\pi}{2}, \pi\right],\; \sin (x)-\cos(x)<1$.
$$\sin(x)<1+\cos(x)$$ $$\sin^2(x)<1+2\cos(x)+\cos^2(x)$$

You should justify this step (squaring an inequality while preserving its direction) by pointing out that $x^2$ is increasing for nonnegative $x$ and $\sin(x)$ is nonnegative on $\left[\frac{\pi}{2}, \pi\right].$

$$1-\cos^2(x)<1+2\cos(x)+\cos^2(x)$$ $$-2\cos^2(x)<2\cos(x)$$
$$-2\cos^2(x)-2\cos(x)<0$$

So far so good.

Dividing by $-2\cos(x)$ leads to:
$$\cos(x)+1<0$$
Since we know that $\cos(x)$ on the interval $\left[\frac{\pi}{2}, \pi\right]$ is
between $0$ and $1$, $\cos(x)+1$ can't be lower than $0$. Thus
contradiction.

Don't you mean that $\cos$ is between $-1$ and $0\,?$ And you neglected to point out that you are dividing by a positive number (because $-2\cos(x)\ge0$ and $x\ne\frac\pi2).$
Alternatively, here's a more straightforward continuation: $$(\cos x)(\cos x+1)>0\\
(\cos x)\;\&\;(\cos x+1)\;\text{are either both positive or both negative}\\
\cos x\notin[-1,0].$$
That is, $\cos x\notin[-1,0]$ for some $x\in\left[\frac{\pi}{2},\pi\right].$
Since, in fact, $\cos x\in[-1,0]$ for all $x\in\left[\frac{\pi}{2}, \pi\right],$ we have a contradiction.
