Why is this easy "proof" of Brouwer's Fixed Point Theorem not correct/common?

In general, $x(\mathbf{u})$ can't be chosen to be continuous. Here is a simple counter-example in $\mathbb R^2$:

$$f(x,y) = \begin{cases} 2xy, & \text{if } y \le \frac12 \\ 1-2(1-x)(1-y), & \text{if } y \ge \frac12 \end{cases}$$

If $y < \frac12$, the unique fixed $x$ is $x=0$; if $y > \frac12$, the unique fixed $x$ is $x=1$. (If $y=\frac12$, then $f(x,y)=x$ fixes all points in $[0,1]$.)