In the following picture, $v,u \in H^1(a,b)$ with $(a,b)\subset \mathbb{R}$.
Well, I think there is a typo.
How do we get $$u(x)+\int^y_x u'(s)ds\leq u(x) +\sqrt{b-a} ||u'||_{L^2(a,b)}\ ?$$
I'm thinking they $\int^y_x u'(s)ds=\int^b_a u'(s)\chi _{(x,y)}ds$, where $\chi _{(x,y)}$ is the indicator function, and we have $||\chi _{(x,y)}||=\sqrt{y-x}\leq \sqrt{b-a}$.
Finally, how do we obtain the 1st inequality involving $u(y)^2$? If we have $f\leq g$, then $f^2\leq g^2$ may not be true if $g<0$. However, we cannot assume that $u$ is positive...
You're right, you should instead do the following with triangle inequality and Holder's inequality for integrals, \begin{align} |u(y)| &\le |u(x)|+ \left| \int_x^y u' \right|\\ &\le |u(x)|+ \int_a^b |u'\chi_{[x,y]}| \\ &\le |u(x)|+ \|\chi_{[x,y]}\|_{L^2[a,b]} \| u'\|_{L^2[a,b]} \\&\leq |u(x)|+ \sqrt{b-a} \|u'\|_{L^2[a,b]}.\end{align} Now you can apply the given elementary inequality and conclude.