Difficulty in understanding a statement in the proof of the correctness of $\text{BFS}$ algorithm as dealt with in CLRS
It is perhaps more illuminating to reformulate this proof by contradiction as a proof by induction. We prove that $d[v] = \delta(s,v)$ by induction on the latter (the case $\delta(s,v) = \infty$ should be taken care of separately). If $\delta(s,v) = 0$ then $v = s$, and indeed $d[s] = 0$. Now suppose that the result holds for all vertices at distance $i$ from $s$, and consider a vertex $v$ at distance $i+1$ from $s$. Let $u$ be a neighbor of $v$ at distance $i$ from $s$. By induction, $d[u] = \delta(s,u)$, and so once $u$ is processed, $d[v] \leq \delta(s,u) + 1 = \delta(s,v)$. Since $d[v] \geq \delta(s,v)$, it follows that $d[v] = \delta(s,v)$, as needed.
How do the two proofs relate? Let $U$ be the set of vertices such that $d[v] = \delta(s,v)$; clearly $s \in U$. Let $v \neq s$ be an arbitrary vertex reachable from $s$, and let $u$ be a predecessor of $v$, that is, a neighbor of $v$ such that $\delta(s,v) = \delta(s,u) + 1$. The proof shows that if $u \in U$ then also $v \in U$. You can now proceed in two different ways:

Proof by contradiction: choose $v \in U$ minimizing $\delta(s,v)$. By definition, the predecessor of $v$ does lie in $U$, and so we reach a contradiction.
Proof by induction: we show that all vertices reachable from $s$ are in $U$, by induction on their distance from $s$.

The two proofs are equivalent. I prefer the second.