Claim: If $K_1$ and $K_2$ are disjoint nonempty compact sets, show that $\exists k_i\in K_i$ such that $0<\vert k_1-k_2\vert=\inf\{\vert x_1-x_2\vert : x_1\in K_1 \quad\land\quad x_2\in K_2\}$.
Can't we simply say that if $0=\vert y_1-y_2\vert=\inf\{\vert x_1-x_2\vert : x_1\in K_1 \quad\land\quad x_2\in K_2\}$, then $y_1\in K_1\cap K_2$ - a contradiction?
here what the book provided:
On
Here's a short proof if you want it:
$K_1$ and $K_2$ are compact, so $K_1 \times K_2$ is compact by Tychonoff. The function $K_1 \times K_2 \to \Bbb R$ sending $(x_1, x_2)$ to $|x_1 - x_2|$ is continuous, so the range is compact, so it has a minimum which can be expressed as $|k_1 - k_2|$ for some $k_1 \in K_1$ and $k_2 \in K_2$.
$A < B = C$ is a shorthand for $A < B$ and $B = C$.
You seem to think that its negation is $A \ge B = C$ (a shorthand for $A \ge B$ and $B = C$)
But that is not the case : its negation is $A \ge B$ or $B \neq C$.
So it is not valid to prove $A < B = C$ by assuming $A \ge B = C$ and getting a contradiction.