Show that if $f\in\mathcal L^1((a,b))$ has a weak derivative, then it has a continuous modification

Question

Let $\lambda$ denote the Lebesgue measure on $\mathcal B(\mathbb R)$, $a,b\in\mathbb R$ with $a<b$, $I:=(a,b)$, $X,Y$ be $\mathbb R$-Banach spaces and $\iota$ be a continuous embedding of $X$ into $Y$.

If $p\in[1,\infty]$, say $f\in\mathcal L^1(I,X)$ has a weak derivative in $L^p(I,Y)$ if there is a $g\in\mathcal L^p(I,Y)$ with $$\int_I\varphi'\iota f\:{\rm d}\lambda=-\int_I\varphi g\:{\rm d}\lambda\;\;\;\text{for all }\varphi\in C_c^\infty(I).$$ In that case, $f':=g$.

I would like to show that

1. if $f\in\mathcal L^1(I,X)$ has a weak derivative in $L^1(I,Y)$, then there is a $\tilde f\in C(I,Y)$ with $\iota f=\tilde f$ $\lambda$-a.e.
2. if $Y=X'$ and $X$ is continuously embedded into a Hilbert space $H$ via $\kappa:X\to H$, then in the situation of (1.) there is a $\tilde f\in C(I,Y)$ with $\kappa f=\tilde f$ $\lambda$-a.e.

Regarding (1.): It's easy to see that if $g\in\mathcal L^1(I,Y)$ and $c\in Y$, then $$f(t):=c+\int_{(a,\:t)}g\:{\rm d}\lambda\;\;\;\text{for }t\in\overline I$$ is continuous and weakly differentiable with $f'=g$.

Now, in the situation of (1.), by the aforementioned fact, $$g(t):=\int_{(a,\:t)}f'\:{\rm d}\lambda\;\;\;\text{for }t\in\overline I$$ is continuous and weakly differentiable with $$g'=f'\tag2.$$

Now, in Mathematical Tools for the Study of the Incompressible Navier-Stokes Equations and Related Models (Corollary II.4.2), the special case $X=Y=\mathbb R$ is considered and we can find the following argumentation

The mentioned lemma is the following:

Maybe I'm missing something, but I don't get why the argumentation is that "complicated". By $(1)$ and $(2)$, it holds $$\int_I\varphi'(\iota f-g)\:{\rm d}\lambda=0\;\;\;\text{for all }\varphi\in C_c^\infty(I)\tag3.$$ Now, if $\varphi\in C_c^\infty(I)$, then $$\psi(t):=\int_a^t\varphi(s)\:{\rm d}s\;\;\;\text{for }t\in I$$ is obviously in $C_c^\infty(I)$ as well and we've got $\psi'=\varphi$. So, we can replace "$\varphi'$" in $(3)$ by "$\varphi$".

And now the du Bois-Reymond lemma should immediately yield $\iota f=g$ $\lambda$-a.e.. Am I missing something?

Regarding (2.): At the moment, I've got no idea why this follows and would need some help to tackle that problem.