Show there is no continuous map $w:\overline{B_1(0)} \to \partial{B_1(0)}$ such that $w(x)=x$ for all $x \in \partial{B_1(0)}$ using Brouwer's fixed point theorem: every continuous map $T:\overline{B_1(0)} \to \overline{B_1(0)}$ has a fixed point in $\overline{B_1(0)}$.
My idea is to argue by contradiction so that there exists a map $w$ which is continuous and all points on the boundary are mapped to itself. Then I need to extend $w$ so that it maps $\overline{B_1(0)} \to \overline{B_1(0)}$ (or I don't need to do this as $\partial{B_1(0)} \subset \overline{B_1(0)}$?) and I can apply Brouwer. Then I can maybe get a jump or something which contradicts continuity somehow?
If there was such a map than consider the line through $x$ and $w(x)$ for each $x$. This line intersects the boundary in some point $y$. Take a map $x \mapsto y$. It would be the map impossible by Brouwer theorem.