How to prove that the set $A = \left\{ {p:{p^2} < 2,p \in {\Bbb Q^+}} \right\}$ has no greatest element?

To show that $A$ has no greatest element, you have to show that there is no $q\in A$ which is greater than any $p\in A$, the opposite of your first line.

If $p^2<2$, we take the number $q=p+\frac{2-p^2}{p+2}=2\frac{p+1}{p+2}$. Then $p<q$ and $q^2=4\frac{(p+1)^2}{(p+2)^2}$. From $p^2<2$ you can infer $\frac{(p+1)^2}{(p+2)^2}<\frac{1}{2}$, so $q^2<2$.

This shows that to every rational $p<2$ we can associate $q=2\frac{p+1}{p+2}$, which will also be rational, and satisfying $p<q$, $q^2<2$, so $p$ is not the largest rational with this property.

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As to where $q=p+\frac{2-p^2}{p+2}$ comes from: We are basically trying to find a root of $t^2-2$, by starting with some non-root $p$ and trying to make a better approximation to $\sqrt{2}$.

Note that the equation $q=p+\frac{2-p^2}{p+2}$ is equivalent to $$(p^2-2)+(p+2)(q-p)=0$$ In other words, $q$ is a root of $t\mapsto (p^2-2)+(p+2)(t-p)$, which is the line which passes through $(p,p^2-2)$ with slope $(p+2)$. If you look at this line, and compare it with the graph of $t\mapsto t^2-2$, you will see that the intersection of the line and the $x$-axis will be closer to $\sqrt{2}$, and still smaller because $p+2>p+\sqrt{2}=\frac{2-p^2}{\sqrt{2}-p}$ (and this is the slope of the secant of the graph of $t^2-2$ passing through $(p,p^2-2)$ and $(\sqrt{2},0)$).

In fact, the slope $(p+2)$ works to make the inequalities in the exercise easier to deal with, but any number greater than $p+\sqrt{2}$ would work as well. Indeed, if $\alpha>p+\sqrt{2}$ and $(p^2-2)+\alpha(q-p)=0$, then $$q=p+\frac{2-p^2}{\alpha}$$ so $q>p$, and $$q<p+\frac{2-p^2}{p+\sqrt{2}}=p+(\sqrt{2}-p)=\sqrt{2}$$ i.e., $q^2<2$. If we choose $\alpha$ rational (so we'd need to know in advance more-or-less how much $\sqrt{2}$ is), then $q$ given as above will also be rational.