Continuity of the supremum of a continuous function over a closed interval Proposition: let $F:R\times [0,1]\rightarrow R$, be a countinuous function. If $g(x):= Sup\{F(x,t):t\in [0,1]\}$, then $g(x)$ is continuous at all points.
My Idea: 
Assume $g$ is not continuous at $x_0$. Then there exists $\epsilon >0$ such that there exists a sequence $\{y_n\}$ which is converges to $x_0$ and $|g(y_n)-g(x_0)|>\epsilon$ and $y_n\ne x_0$ for all $n$ .
$g(y_n)=F(y_n,t_n)$ for some $t_n\in [0,1]$ as $[0,1]$ is compact and $F$ is continuous. Thus we have the sequence $\{(y_n,t_n)\}$. Now since the sequence lies in a closed interval by the Bolzano–Weierstrass theorem there is a sub-sequence $\{(y_m,t_m)\}$ which converges. It is easy to note $\{(y_m,t_m)\}$ converges to $(x_0,t_0)$ where $t_0$ is some real number between 0 and 1 .
Hence $F(y_m,t_m)>g(x_0)+\epsilon$ or $F(y_m,t_m)<g(x_0)-\epsilon$
as 
$g(y_m)>g(x_0)+\epsilon$ or $g(y_m)<g(x_0)-\epsilon$, which implies 
$F(x_0,t_0)\ge g(x_0)+\epsilon$ or $F(x_0,t_0)\le g(x_0)-\epsilon$ 
If it were only for the first inequality we would be done, alas it is not the case.
I would appreciate if someone helped me prove/disprove the proposition.
 A: We show $g(x)$ is continuous by the $\epsilon- \delta$ definition using uniform continuity of $F(x)$.
To show $g(x)$ is continuous at $x_0$, given $\epsilon>0$ we note that $F(x)$ is uniformly continuous in the interval $[x_0-1\ ,\ x_0+1]\times [0,1]$.
Next given ${\epsilon}$ by uniform continuity, $\exists\ \delta>0$ such that $|F(\textbf{x})-F(\textbf{y})|<\epsilon \ \ \ \forall \ \textbf{x,y}\in[x_0-1\ ,\ x_0+1]\times [0,1]\ \ \ \&\ \ \ ||\textbf{x-y}||<\delta$
Now given $x\in \ (x_0-min(\delta,1),\ x_0+min(\delta,1))$


*

*Case 1: $g(x)> g(x_0)+\epsilon$ : let $F(x,t)=g(x)$ we know such a $t$ exists as $F$ is continuous. Note $|F(x,t)-F(x_0,t)|<\epsilon$ or $F(x_0,t)>g(x)-\epsilon>g(x_0)$  which is a contradiction as $g(x_0)\ge F(x_0,a)\ \forall a\in[0,1]$

*Case 2: $g(x)< g(x_0)-\epsilon$ : let $F(x_0,t)=g(x_0)$ we know such a $t$ exists as $F$ is continuous. Note $|F(x,t)-F(x_0,t)|<\epsilon$ or $F(x,t)>g(x_0)-\epsilon>g(x)$  which is a contradiction as $g(x)\ge F(x,a)\ \forall a\in[0,1]$
what remains is $|g(x)-g(x_0)|\le \epsilon$
Hence $g(x)$ is continuous everywhere!!
