There exists a constant $C > 0$ such that
$$\int_a^b |f(x)|^2 \,dx \leq C \int_a^b |f'(x)|^2 \, dx$$
for all $f \in C^1([a,b])$ such that $f(b) = 0$
What I have attempted is to say that
$$|f(b) - f(x)| = \int_x^b |f'(y)| \, dy \\ \leq \int_x^b|1^2|^{0.5} \, dy \cdot \int_x^b |f'(y)^2|^{0.5} \, dy \\ = |b-x|^{0.5} \cdot \int_x^b |f'(y)^2|^{0.5} \, dy$$
Then by squaring and integrating both sides I get
$$\int_a^b |f(b)-f(x)|^2 \, dx \leq \int_a^b (b-a) \, dx \int_x^b f'(y) \, dy$$
But this is where I am stuck. I don't think I can say that this is
$$\leq (b-a)^2 \int_a^b f'(y) \, dy$$
A second thought on this is that the question I have been asked is incorrect and it should have Boundary conditions that say $f(a)=f(b)=0$, but I have no way to know whether this is true or not.
I think the way your argument starts can lead to a solution. Indeed,
since $f(b) = 0$ we have $-f(x) = \int_x^b f'(y) dy$, for all $x\in [a,b]$. Hence $$ |f(x)| \leq \int_x^b |f'(y)| dy \leq (b-x)^{1/2} \left(\int_{x}^b|f'(y) |^2 dy \right)^{1/2}, $$ where we have used Hölder's inequality. Now, squaring both sides and integrating over $x\in [a,b]$ we get $$ \int_a^b |f(x)|^2 dx \leq \int_a^b(b-x)\int_x^b|f'(y)|^2dydx \leq \int_a^b |f'(y)|^2 dy\int_a^b(b-x)dx = \\ \frac{(b-a)^2}{2} \int_a^b |f'(y)|^2 dy, $$ which is the desired inequality with $C = \frac{(b-a)^2}{2}$.