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Let $V$ be real normed vector space of Riemann integrable functions $f : [0,1]tomathbbR$ with norm

$||f||:=sup:xin[0,1]$.

Let $f_n:[0,1]tomathbbR$ be continuous for $ninmathbbN$ and let there be a function $fin V$ such that

$lim_ntoinfty ||f-f_n|| = 0$.

I have to show that $lim_ntoinfty int_0^1f_n(x)dx=int^1_0f(x)dx$.

I have already shown that $f$ is continous.

My strategy:

Prove that $lim_ntoinfty int_0^1f_n(x)dx -int^1_0f(x)dx = 0$

so $lim_ntoinfty int_0^1(f_n(x) - f(x))dx = 0$.

I then think I have to use the squeeze theorem:

$lim_ntoinftyinf_[0,1](f_n(x) - f(x)) leqlim_ntoinfty int_0^1(f_n(x) - f(x))dx leq lim_ntoinftysup_[0,1](f_n(x) - f(x))$

Is it then allowed to put the limit inside the supremum and infimum, which makes them both equal to 0?

You are almost done, just modify the last line as
$$0leq left|int_0^1(f_n(x) - f(x))dxright|leq int_0^1left|f_n(x) - f(x)right|dx leq sup_[0,1]left|f_n(x) - f(x)right|$$
then use $sup_[0,1]left|f_n(x) - f(x)right|=|f-f_n|to 0$
and apply the squeeze theorem.

P.S. Following your approach, you need
$$-sup_[0,1]left|f_n(x) - f(x)right|leqinf_[0,1](f_n(x) - f(x)) leq int_0^1(f_n(x) - f(x))dx\ leq sup_[0,1](f_n(x) - f(x))leqsup_[0,1]left|f_n(x) - f(x)right|$$
and again use $sup_[0,1]left|f_n(x) - f(x)right|=|f-f_n|to 0$.

Is it then allowed to put the limit inside the supremum and infimum, which makes them both equal to 0?

You can't switch infimum/supremum and the limit in general.

Consider $f_n(x)=maxx-n$. Then $sup_xinmathbb R|f_n(x)|=1$ while $lim_ntoinfty |f_n(x)|=0$. Hence,
$$
lim_ntoinftysup_xinmathbb R|f_n(x)|=1neq 0=sup_xinmathbb Rlim_ntoinfty |f_n(x)|.
$$

But your case is a bit more special because $[0,1]$ is compact and $f_n$ and $f$ are continuous. Hence, the infimum and supremum are achieved at some points. Then you can apply the limit as your last step.

The map $f mapsto int_0^1 f(x),dx$ is continuous w.r.t your norm $|cdot|$.

Indeed, it is linear and bounded:

$$left|int_0^1 f(x),dxright| le int_0^1 |f(x)|,dx le int_0^1 |f|,dx = |f|$$

Therefore $f_n xrightarrow f$ implies $ int_0^1 f_n(x),dx to int_0^1 f(x),dx$