The definition of $(\xi,\xi')_o$ is an infimum taken over all pairs $x$, $x'$, such that $x=(x_i)$ is a sequence representing $\xi$, and $x'=(x'_i)$ is a sequence representing $\xi'$. The quantity being infimized is $\liminf_{i \to \infty} (x_i,x'_i)_o$.
For any nonempty set of real numbers that has an infimum, there is a sequence in that set such that the limit of the sequence equals the infimum of the set.
Therefore, there exists a sequence of pairs $x^n$, $x'^n$ such that each $x^n=(x^n_i)$ is a sequence representing $\xi$, each $x'{}^n = (x'{}^n_i)$ is a sequence representing $\xi'$, and
$$(\xi,\xi')_o = \lim_{n \to \infty} \left( \liminf_{i \to \infty} \underbrace{(x^n_i,x'{}^n_i)_o}_{G(n,i)} \right)
$$
Here's where the "diagonalization" procedure kicks in. Imagine the doubly indexed sequence of real numbers $G(n,i)=(x^n_i,x'{}^n_i)_o$ arranged in a table, with $n$ indexing the rows and $i$ indexing the columns. Knowing the above equation, you carefully choose a sequence of entries $G(n,i_n)$ out of the table so that $(\xi,\xi')_o = \lim_{n \to \infty} G(n,i_n)$:
- Choose $i_1$ so that the row 1 entry $G(1,i_1)$ is within distance $2^{-1}$ of the liminf of the row $1$;
- Choose $i_2$ so that the row 2 entry $G(2,i_2)$ is within distance $2^{-1}$ of the liminf of row $2$;
- Choose $i_3$ so that the row $3$ entry $G(3,i_3)$ is within distance $2^{-3}$ of the liminf of row $3$;
.
.
.
The sequence $G(n,i_n)$ converges to $(\xi,\xi')_0$ by the sandwich lemma:
$$\liminf_{i \to \infty} (x^n_i,x'{}^n_i) - 2^{-n} \le G(n,i_n) \le \liminf_{i \to \infty} (x^n_i,x'{}^n_i) + 2^{-n}
$$
and since $\liminf_{i \to \infty} (x^n_i,x'{}^n_i)$ converges to $(\xi,\xi')_o$ and $2^{-n}$ converges to $0$, it follows that $G(n,i_n)$ converges to $(\xi,\xi')_o$.
