I have to prove that there is no function
$$f: B[0,1] \to \mathbb{S}^1 $$
such that
$$\left.f\right |_{\partial B[0, 1]} = id \tag{$\star$}$$
I want to use the concept of fundamental groups (however if one has a another idea please feel free to share).
I tried to assume that there exists such a function $f: B[0,1] \to \mathbb{S}^1$ satisfying $(\star)$. Then the identity map on the boundary induces the trivial homomorphism on the fundamental groups $f_{*} : \pi_1(\partial B[0, 1]) \to \pi_1 (\mathbb{S}^1)$. Also $ \pi_1(\partial B[0, 1]) = \pi_1(\mathbb{S}^1) $ because the boundary of the closed unit ball is homeomorphic to the unit circle.
Is this argument enough? Thank you in advance.
I don't exactly follow your argument. If you take the identity map, then it should induce an isomorphism of fundamental groups.
I hope this is known: if $A$ is a retract of $X$, then $\pi_1(A,X) \to \pi_1(X, x_0)$ is a monomorphism for each $x_0 \in A$. If not, this follows easily from functoriality of the fundamental group.
If $S^1$ were a retract of $B^2$, then its fundamental group would be a subgroup, but $\pi_1(S^1, \cdot) = \mathbb{Z}$ and $\pi_1(B^2, \cdot) = \{ 1 \}$.
This statement is equivalent to Brouwer's theorem, which states that every map $f: B^2 \to B^2$ has a fixed point. It has lots of known proofs, available all over the internet, if you are interested.