Why is a lower bound necessary in proofs of VC-dimensions for various examples of hypotheses?
Lets use the quotes from the book.


  To give an upper bound, we need to prove that no set $S$ of cardinality
  $d + 1$ can be shattered by $H$


By doing this, we are proving that $\text{VCdim}(H) &lt; d+1$, but it does not necessarily mean $\text{VCdim}(H) = d$. For example, we may go for an easier-to-prove failure like sets with $2d+1$ members, and consequently prove $\text{VCdim}(H) &lt; 2d+1$, although we know $\text{VCdim}(H) = d$. Therefore:


  To give a lower bound $d$ for $\text{VCdim}(H)$, it suffices to show that a set $S$
  of cardinality $d$ can be shattered by $H$


We need to prove the lower bound as well to show that $d \le \text{VCdim}(H) &lt; d+1$, which implies $\text{VCdim}(H) = d$. The last equality means that for each size $1$ to $d$, $H$ can shatter at least one set $S$ with that size.

Note that for the $2d+1$ example, we will fail to prove $2d \le \text{VCdim}(H)$, therefore $\text{VCdim}(H) \neq 2d$, and consequently we must try to prove a smaller (probably harder to prove) upper bound $d+1$.