k×k grid has tree-width at least k I am looking for ideas how to solve the problem from Diestel's textbook Graph Theory.
Chapter 12. Minors, Trees, and WQO.
Problem 16. Apply Theorem 12.3.9 to show that the $k \times k$ grid has tree-width at least $k$, and find a tree-decomposition of width exactly $k$.

Theorem 12.3.9. (Seymour & Thomas 1993). Let $k \geq 0$ be an integer. A graph has tree-width $\geq k$ if and only if it contains a bramble of order $> k$. (A bramble is a set of mutually touching connected vertex sets in $G$; its order is the minimum number of vertices in a set meeting each member of the bramble.)

The problem is the proof of the theorem 12.3.9 was given in the terms of bramble, which is a bit confusing, at present I don't really see the way to solve the problem by using this theorem. 
If you familiar with the topic, please, help me out.
Addendum:
In Graphs & Algorithms: Advanced Topics on the slide 5.
The $n\times n$ - grid on $\left \{(i,k) | 1 \leq i, j \leq n\right \}$ has treewidth $\leq n$: Consider the path on
$X_{n(i-1)+j}=\left \{(i,k)|j\leq k\leq n\right \}\cup\left \{(i+1,k)|1\leq k\leq j\right \}, 1\leq i\leq n-1, 1\leq j\leq n$
How this is supposed to help me?

I didn't read Diestel's book, but simple observation may be help is, if $G$ is a grid then we know $\operatorname{tw}(G) \ge BN(G)-1$ and as you know if you select $row_i \cup col_j$ as a bramble set, its hitting set size is $n$, so treewidth is at least $n-1$. Also you can simply construct a tree decomposition with size of $n$. So treewidth is $n$ or $n-1$. (By $BN(G)$ I mean bramble number of $G$.)
Edit: As Braian mentioned you can simply find good brambles and answer of sample tree decomposition is in your question. (my mistake was I thinking about $cross_{i,i}$ not $cross_{i,j}$
