analysis - Let $f$ be continuous real-valued function on $[0,1]$. Then, $F(x)=max{f(t):0leq tleq x}$ is continuous

Let $f$ be continuous real-valued function on $[0,1]$ and

$$\begin{align} F(x)=\max\{f(t):0\leq t\leq x\}. \end{align}$$
I want to show that $F(x)$ is also continuous on $[0,1]$.

MY WORK

Let $\epsilon> 0$ be given and $x_0\in [0,1].$ Since f is continuous at $x_0\in [0,1],$ then $\forall x\in [0,1]$ with $|x-x_0|<\delta,$ it implies $|f(x)-f(x_0)|<\epsilon.$

Also, $$\begin{align} |f(t)-f(x_0)|<\epsilon, \text{whenever}\; |t-x_0|<\delta,\;\forall\; t\in[0,x]\end{align}$$

Taking max over $t\in[0,x]$, we have
$$\begin{align} \max|f(t)-f(x_0)|<\epsilon, \text{whenever}\; |t-x_0|<\delta,\;\forall\; t\in[0,x]\end{align}$$
$$\begin{align} |\max f(t)-\max f(x_0)|<\epsilon, \text{whenever}\; \max|t-x_0|<\delta,\;\forall\; t\in[0,x]\end{align}$$
$$\begin{align} |F(x)-f(x_0)|<\epsilon, \text{whenever}\; |x-x_0|<\delta\end{align}$$

which implies that $F(x)$ is continuous on $[0,1].$

I am very skeptical about this proof of mine. Please, is this proof correct? If no, a better proof is desired. Thanks!

Answer

Your proof is wring because you cannot take maximum over $t \in [0,x]$ in an inequality which is valid only for $|x-t| <\delta$. Here are some hints for a correct proof. Verify that $|F(x)-F(y)| \leq \max \{|f(t)-f(s)|:x\leq t \leq y, x\leq t \leq y\}$ for $x