Solution. Continuity at \(c\): given \(\varepsilon >0\), for \(\left |x-c\right |<\delta \leq 1\), \[\left |x^{2}-c^{2}\right | = \left |x-c\right |\left |x+c\right | \leq \left |x-c\right |\left (2\left |c\right |+1\right ),\] so \(\delta = \min \left \{1,\ \varepsilon /(2\left |c\right |+1)\right \}\) works.

That \(\delta \) shrinks as \(\left |c\right |\) grows, and no single choice serves. Take \(x_n = n + \tfrac 1n\) and \(y_n = n\). Then \(\left |x_n-y_n\right | = \tfrac 1n\rightarrow 0\), yet \[\left |f(x_n)-f(y_n)\right | = \left (n+\tfrac 1n\right )^{2}-n^{2} = 2 + \frac {1}{n^{2}} > 2 .\] So no \(\delta \) works for \(\varepsilon = 2\), and \(f\) is not uniformly continuous.

On any bounded interval it is uniformly continuous, by the theorem above. The failure is entirely about the unbounded domain.

Solution. Define \(g(x) = f(x)-x\), continuous on \([0,1]\) as a difference of continuous functions. Since \(f\) maps into \([0,1]\), \[g(0) = f(0)-0 = f(0) \geq 0,\qquad g(1) = f(1)-1 \leq 0 .\] If either is zero the corresponding endpoint is a fixed point. Otherwise \(g(0)>0>g(1)\), and by the intermediate value theorem \(g(c)=0\) for some \(c\in (0,1)\), that is \(f(c)=c\).

The argument uses only continuity and the fact that the interval maps into itself. It fails on \((0,1)\): the map \(x\mapsto x/2\) is continuous and maps \((0,1)\) into itself with no fixed point in it, the fixed point \(0\) having been excluded with the endpoint.