Continuity of a function in $\mathbb{R}$ Suppose $f\colon [a, b] \to \Bbb R$ is continuous. Define $F(x):= \sup \{f(y)\mid y \in [a, x]\}$. Prove $F(x)$ is continuous. 
What I have so far:
For $x \le b$, the set $\sup\{f(x)\mid y \in [a, x]\} \not= \emptyset$. By Completeness then, $\sup\{f(y)\mid y \in [a, x]\}$ exists and is finite. So by the Intermediate Value Theorem, there exists a $x_0\in [a, x]$ such that $f(x_0) = s_{0_k}$, where $s_{0_k} = \sup\{f(x)\mid y \in [a, x]\}$.
Then choose a sequence $\{s_n\}$ in $[a, x]$ such that $s_n\to s_{0_k}$. Since there exists a sequence in a closed, bounded set that converges to the supremum of that set.
Also then $s_{0_k}\in [a, x]$ by the Comparison Theorem. And since $f$ is continuous, $\lim f(s_n) = s_{0_k}$. 
Observe that as $x_n\to b$ as $n \to \infty$ and that $s_{0_k}$ is monotone increasing. Thus each $s_{0_k}\to s_0$ where $s_0=\sup\{f(y)\mid y \in [a, x]\}$. Thus each $s_{0_k}\to s_0$, hence $F(x)$ is continuous at each $s_{0_k} \in [a, b]$.
First off, is this correct? Secondly, It seems to me there is a simple answer, but I can't see it any ideas? 
 A: This seems correct. As for a simpler answer, show that $F$ is increasing on $[a,b]$ then show that if $F$ is not continuous at some $x \in (a,b]$, then $F(x) = f(x)$ and
$$F(x) > F(x^-)  \Longrightarrow  f(x) > \sup_{y \in [a,x)} f(y) \geq \lim_{y \to x^-} f(y)    $$
which contradicts that $f$ is continuous.
A: Hint. For any $x_1,x_2\in[a,b]$, with $x_1<x_2$:
$$
F(x_2)=\sup_{x\in[a,x_2]} f(x)=\max\Big\{\sup_{x\in[a,x_1]} f(x),\sup_{x\in[x_1,x_2]} f(x) \Big\}.
$$
Let $x_0\in[a,b]$ and $\varepsilon>0$. Since $f$ is continuous at $x_0$, there is a $\delta>0$, such that: $|x-x_0|<\delta\Leftarrow |f(x)-f(x_0)|<\varepsilon$ or equivalently
$$
|x-x_0|<\delta\quad\Longrightarrow\quad f(x_0)-\varepsilon<f(x)<f(x_0)+\varepsilon.
$$
Hence for every $x\in (x_0,x_0+\delta)$:
\begin{align}
F(x) &=\max\Big\{\sup_{x\in[a,x_0]} f(x),\sup_{x\in[x_0,x]} f(x) \Big\}\le 
\max\Big\{\sup_{x\in[a,x_0]} f(x),\sup_{x\in[x_0,x_0+\delta)} f(x) \Big\} \\ &\le
\max\Big\{\sup_{x\in[a,x_0]} f(x),\,f(x_0)+\varepsilon \Big\}\le \varepsilon+\sup_{x\in[a,x_0]} f(x)=F(x)+\varepsilon.
\end{align}
