I was going through section of Breadth First Search of the text Introduction to Algorithms by Cormen et. al. and I faced difficulty in understanding a statement in the proof below which I have marked with $\dagger$
Theorem: (Correctness of breadth-first search)
Let $G = (V, E)$ be a directed or undirected graph, and suppose that $BFS$ is run on $G$ from a given source vertex $s \in V$. Then, during its execution, $BFS$ discovers every vertex $v \in V$ that is reachable from the source $s$, and upon termination, $d[v] = \delta(s, v)$ for all $v \in V$. Moreover, for any vertex $v \neq s$ that is reachable from $s$, one of the shortest paths from $s$ to $v$ is a shortest path from $s$ to $\pi[v]$ followed by the edge $(\pi[v], v)$.
$\left[ \delta(s,v) \rightarrow \text{Length of the shortest path from s to v}\\ d[v]\rightarrow \text{distance assigned to vertex $v$ from $s$ by BFS}\\\pi[v]\rightarrow \text{Predecessor of $v$ in the path from $s$ to $v$ in the BFS}\right]$
Proof:
Assume, for the purpose of contradiction, that some vertex receives a $d$ value not equal to its shortest path distance. Let $v$ be the vertex with minimum $\delta(s, v)$ that receives such an incorrect $d$ value; clearly $v \neq s$. We know $d[v] \geq \delta(s, v)$, and thus we have that $d[v] > \delta(s, v)$. Vertex $v$ must be reachable from $s$, for if it is not, then $\delta(s, v) = \infty \geq d[v]$. Let $u$ be the vertex immediately preceding $v$ on a shortest path from $s$ to $v$, so that $\delta(s, v) = \delta(s, u) + 1$. Because $\delta(s, u) < \delta(s, v)$, and because of how we chose $v$ , we have $d[u] = \delta(s, u)$$^\dagger$. Putting these properties together, we have
$d[v] > \delta(s, v) = \delta(s, u) + 1 = d[u] + 1 $
(The proof then continues, but I do not include the rest here...)
I could not understand the reasoning behind the statement marked with $\dagger$, $\text{"because of how we chose $v$ , we have $d[u] = \delta(s, u)$"}$