Suppose $E\subseteq \mathbb{R}$. A function $f: E\to\mathbb{R}$ is said to be upper semicontinuous at $x_0 \in E$ if for any $\varepsilon>0$, there exists $\delta>0$ such that for any $x\in E$, $|x-x_0|<\delta$ then $f(x)<f(x_0)+\varepsilon$. If $f$ is upper semicontinuous at every point of $E$, then $f$ is upper semicontinuous on $E$.
Let $u: \mathbb{R}\to [0,1]$ be a function defined by \begin{align*} u(x)= \begin{cases} \frac{1}{2}, \quad x\leq 0,\\ \frac{x}{2},\quad 0<x<1,\\ 1,\quad x\geq 1. \end{cases} \end{align*}
I want to show that $u$ is upper semicontinuous on $\mathbb{R}$. Since continuous function is both upper semicontinuous and lower semicontinuous, so I think, I just need to show that $u$ is upper semicontinuous at its discontinuity points, i.e $0$ and $1$.
For $x=0$ (then $u(x)=\frac{1}{2}$). Let $\varepsilon>0$. Choose $\delta=\frac{1}{2}$. Take any $y\in\mathbb{R}$ such that $|y-0|=|y|<\frac{1}{2}$ . Therefore $-\frac{1}{2}<y<\frac{1}{2}$. Then, for $-\frac{1}{2}<y<0$, we have $u(y)=\frac{1}{2}<\frac{1}{2}+\varepsilon.$ For $0<y<\frac{1}{2}$, we have $u(y)<\frac{1}{4}<\frac{1}{2}+\varepsilon.$ So, $u$ is upper semicontinuous at $x=0$.
For $x=1$ (then $u(x)=1$). Let $\varepsilon>0$. Choose $\delta=\frac{1}{2}$. Take any $y\in\mathbb{R}$ such that $|y-1|<\frac{1}{2}$ . Therefore $\frac{1}{2}<y<\frac{3}{2}$. Then, for $\frac{1}{2}<y<1$, we have $u(y)<\frac{1}{2}<1+\varepsilon.$ For $1\leq y<\frac{3}{2}$, we have $u(y)=1<1+\varepsilon.$ So, $u$ is upper semicontinuous at $x=1$.
Is it correct? Thanks for any help.