So let me first state the problem and then I'll ask my question. Let $I$ be the unit interval and $q\geq p\geq 1$. Show that there exists a constant $C=C(p,q,I)$ such that
$$||u||_{L^q(I)}\leq C||u||_{W^{1,p}(I)}$$
for all $u\in W_0^{1,p}(I)$.
So here is my question. Shouldn't $q$ and $p$ be in the other order. In other words $p\geq q \geq 1$ instead of $q\geq p\geq 1$. I have copied the statement of the problem verbatim, but it seems unsolvable in the original form. I am wondering if it is a possible typo. Now if we assume that $p\geq q\geq 1$ I can solve it no problem. I have included the proof below
If we assume $p\geq q \geq 1$ then I can prove it included below.From a density argument we may assume that $u\in C_c^\infty$. It suffices to show that $||u||_{L^q(I)}\leq C_1||u||_{L^p(I)}$ and $||u||_{L^q(I)}\leq C_2||Du||_{L^{p}(I)}$. Then we can sum the inequalities and set $C=\max{\{C_1,C_2\}}$. Since $p/q\geq 1$ we have by holder's inequality
$$\begin{align} \int_I |u|^qdx &\leq (\int_I|u|^pdx)^{q/p} \end{align}$$
Thus, $||u||_{L^q}\leq ||u||_{L^p}$.Furhtermore,
$$\begin{align} |u(x)| &= |\int_0^xDu(t)dt|\\ &\leq\int_0^1|Du|^pdt \end{align}$$
The second inequality coming from Holder's ienquality. Taking the $L^q$ norm on both sides we get $||u||_{L^q(I)}\leq ||Du||_{L^p(I)}$ and were done.
The problem as stated is correct. Your first step that bounds the q-norm by the p-norm misleads you. You need to trade the derivative for integrability. Otherwise what you are doing is just a Poincare-type inequality, not Sobolev-type.