$X,Y$ spaces such that are cones over $X\cap Y$ generate $X\cup Y=\sum (X\cap Y)$

I have this question from this Kratzer and Thevenaz article :

In page 89, they explain that

If $ Y=Y_1\cup Y_2$ are such that $ X=Y_1\cap Y_2$ and if each $ Y_i$ is a cone over $ X$ , then $ Y$ is the suspension of $ X$

But in page 90, in the proof of the Proposition 2.5, they use this in a way I don’t understand. They prove that $ \mid X\mid \simeq C\mid E\mid $ and $ \mid Y\mid \simeq C\mid F\mid$ , but those aren’t cones over $ \mid X\cap Y\mid$ , so why I can continue the proof of this proposition saying that this implications make $ \mid G\mid \simeq \sum(\mid X\cap Y\mid)$ ?

My concept of a “cone over $ \mid X\cap Y\mid$ ” is:

$ $ \mid X\cap Y\mid\times [0,1]/ (a,i)\thicksim(b,1)$ $

with $ a,b\in X\cap Y$ .

Am I misunderstanding what is a cone over $ \mid X\cap Y\mid$ ?