Continuous function in [0,1] for each x f(x)>x

Question

Continuous function in [0,1] for each x f(x)>x, Prove that exists c>0 for each x in [0,1] f(x)>x+c.

Answer 1

$g(x)=f(x)-x$ is continious and positive in [0,1], so has minimum $g(c_0)=a>0$. Take $c=\frac{a}{2}$

Answer 2

Assume the contrary; then there exists a sequence $(x_n)_n$ in $[0,1]$ such that $|f(x_n)-x_n|\rightarrow0$. By Bolzano-Weierstrass, $(x_n)_n$ has a convergent subsequence $(x_{n_k})_k$ with limit $x\in[0,1]$. By continuity, $f(x_{n_k})\rightarrow f(x)$. But then $|f(x_{n_k})-x_{n_k}|\rightarrow|f(x)-x|$, so we have $f(x)=x$; contradiction.