Suppose that $f \colon [0, 1] \to \mathbb{R}$ is infinitely continuously differentiable and satisfies $f(0) = f(1) = 0$.
We define two (semi-)norms, $$ \|f\|_{\rm Lip} := \sup_{x \neq y} \frac{|f(x) - f(y)|}{|x - y|}. $$ Also define $$ \|f\|_{W}^2 = \int_0^1 (f')^2(x) \, dx. $$ Evidently $\|f\|_{W} \leq \|f\|_{\rm Lip}$. However, is there a converse inequality for the set-up described above of the form $\|f\|_{\rm Lip} \leq C \|f\|_{W^{1, 2}_0}$ which holds for all such $f$ for some constant $C \geq 1$?
No, the best you can do is Morrey's inequality which replaces $Lip$ with $C^{1/2}$. Here is a simple example: consider the seminorms $$[f]_{C^\alpha}=\sup_{x\neq y}\frac{|f(x)-f(y)|}{|x-y|^\alpha}.$$ Fix $\epsilon>0$ small and let $f$ be a function that equals $|x|^{- \beta}$ when $x>\epsilon$ but gets smoothly rounded off to a constant when $0\leq x<\epsilon$. We can estimate the orders of magnitude $$[f]_{C^\alpha}\sim \frac{\epsilon^{-\beta}}{\epsilon^\alpha}=\epsilon^{-\alpha-\beta}.$$ On the other hand, $$\|f\|_{W^{1,2}}^2\sim\int_{\epsilon}^1|x|^{2(-\beta-1)}dx\sim\epsilon^{-2\beta-1}.$$ Thus the claim $[f]_{C^\alpha}\lesssim[f]_{W^{1,2}}$ in this case is equivalent to $$\epsilon^{-\alpha-\beta}\lesssim\epsilon^{-\beta-1/2}.$$ Taking $\epsilon$ small, we need $$\alpha\leq\frac12.$$ Note that your question pertains to the $\alpha=1$ case.