How to prove $f:\mathbb R→[0,1]$ with $f(x)=0$ for $x\leq 0$ and $f(x)=x$ for $x\in[0,1]$, $f(x)=1$ for $x\geq 1$ is not an open mapping?

Question

Let $f:\mathbb{R}\rightarrow [0,1]$ be defined as$$ f(x) = \begin{cases} 0; & x \le 0\\ x; & x \in [0,1]\\ 1; & x \ge 1 \end{cases} $$ Why it is not an open mapping?

Answer 1

Because $U=(-\infty, 0)\subseteq \mathbb{R}$ is open, but $f(U)=\{0\}\subseteq [0,1]$ is not open.