We have $4$ points: $P_1 =[0:0:0:1], P_2=[1:0:0:-1], P_3=[1:0:0:1], P_4=[1:0:0:0]$ in $\mathbb{P}^3$. I want to show that this set of points is cross- ratio.
I am trying to prove it by use equation: $$(P_1,P_2,P_3,P_4) = \frac{(P_3-P_1)(P_4-P_2)}{(P_3-P_2)(P_4-P_1)}.$$
I obtain that: $$(P_1,P_2,P_3,P_4) = \frac{([1:0:0:1]-[0:0:0:1])([1:0:0:0]-[1:0:0:-1])}{([1:0:0:1]-[1:0:0:-1])([1:0:0:0]-[0:0:0:1])}= \frac{[1:0:0:0]\cdot [0:0:0:1]}{2[0:0:0:1]\cdot [1:0:0:-1]}.$$ And I know that it should be equal $-1$ or $\frac{1}{2}$, when we want to cross-ratio. But how I should prove it?
There seems to be some serious misunderstandings here. The cross-ratio is a value assigned to 4 points on a line (or 4 projective points on a projective line). You cannot say that "4 points are cross-ratio" as that does not make sense. Instead you would say "the cross ratio of 4 collinear points is ...". In particular you seem to be looking for the statement that your 4 points have cross-ratio $-1$ (the harmonic cross-ratio) - reordering your points will switch this $-1$ to $2$ or $\frac{1}{2}$ so we can include those as possibilities as well.
So firstly we must check your 4 points are collinear. This is straightforward as they clearly lie on the projectivisation of the plane spanned by $(1,0,0,0)$ and $(0,0,0,1)$ so they all lie on this projective line. Now to calculate the cross ratio we should take your representatives and check if they lie on an affine line in $\mathbb{R}^4$. In fact they do not but lets look at what we would do if they did first:
With 4 collinear points in $\mathbb{R}^4$, $v_1, v_2, v_3, v_4$ we calculate their cross-ratio as (note the modulus signs): $$ (v_1, v_2; v_3, v_4):= \frac{|v_1-v_3||v_2-v_4|}{|v_1-v_4||v_2-v_3|} \in \mathbb{R} \cup \{\infty\}$$.
Now back to the situation in hand. Let us denote by $x_i$ the representative of $P_i$ so e.g. $x_2 = (1,0,0,-1)$. 3 of our representatives lie on a line but $x_1$ is parallel to that line. In other words it represents the point at infinity. Now we can't just plug infinity into our equation but if we take the limit of the cross-ratio as one of the points tends to infinity we see that the terms with it in cancel so we get:
$$ (P_1, P_2; P_3, P_4) = (x_1, x_2; x_3, x_4) = \frac{|x_2-x_4|}{|x_2-x_3|} = \frac{1}{2}$$
If we swap the roles of $x_2, x_4$ we get $(x_1, x_4; x_3, x_2) = -1$