Proof verification for "continuity of function $f\implies $ continuity of $m( x) =\min_{a\leq t\leq x} f( t)$". Let $\displaystyle f:[ a,b]\rightarrow \mathbb{R}$ be a continuous function, that is $\displaystyle f\in \mathcal{C}[ a,b]$. It is to be proven that $\displaystyle m\in \mathcal{C}[ a,b]$, where $\displaystyle m( x) =\min_{a\leq t\leq x} \ f( t)$.
I divide it into two parts:
First part: It can be shown that $\displaystyle m$ is monotonically decreasing.
Since $\displaystyle m$ is monotonic, its one sided limits exist. Idea is to show there is no discontinuity (jump discontinuity) at any $\displaystyle c\in [ a,b]$ and this is achieved by showing $\displaystyle m( c-) =m( c+)$.
Second part:
Fix any ${\displaystyle \delta  >0}$,
\begin{align}
0\leq m(c-)-m(c+)=&\inf \{m(x):a\leq x< c\}-\sup \{m(y):c< y\leq b\}\\
 =&\inf \{m(x):c-\delta < x< c\}-\sup \{m(y):c< y< c+\delta \}\\ =&\inf \{m(x)-m(y):c-\delta < x< c< y< c+\delta \}\\
\Longrightarrow &0\le m(c-)-m(c+)\leq m(x)-m(y) \tag1
\end{align}
Here in $(1)$,  $x$ is any point in $(c-\delta,c)$ and $y$ is any point in $(c,c+\delta)$.
By continuity of $\displaystyle f$, minimum value is attained by $\displaystyle f$. In particular, $m(y)$ is attained by $f$ on $[a,y\rbrack$. Let's call the point $z$ where minimum value of $f$ on $[a,y\rbrack$ that is $m(y)$ is attained. There are two cases: Case 1): $z\le x$, Case 2): $y\ge z\gt x$. 
Case 1):  It follows that $m(y)=\min_{a\le t \le y} f(t)=f(z)$ for some $z\leq x$, whence it follows that $m(x)=f(z)$ (By definition of $m$).
It follows from ${\displaystyle (1)}$ that ${\displaystyle m(c-)=m(c+)}$ and we are done! Therefore, we have shown that from $z\le x$ it follows that $m(c-)=m(c+)$ thereby proving continuity at $c$.
Case 2): $y\ge z\gt x$ 
That is, $m(y)$ is attained by $f$ at $z\gt x$ that is ${\displaystyle m(y)=f(z)}$. Here the idea is to bring $t$ (this variable $t$ is introduced in $(2)$) and $z$ in some delta neighborhood of $c$ and this will help us apply uniform continuity later.
From ${\displaystyle (1)}$, we get: \begin{equation}
0\leq m(c-)-m(c+)\leq m(x)-f(z)\le f(t)-f(z) \tag{2}
\end{equation}
Here ${\displaystyle t\in (c-\delta ,x)}$. Clearly right hand side on ${\displaystyle (2)}$ is positive.
By uniform continuity of ${\displaystyle f\ }$ on $\displaystyle [ a,b]$, for any ${\displaystyle \epsilon  >0,}$ there exists a ${\displaystyle \delta _{1}  >0}$, such that $ $for all ${\displaystyle p,q}$ in ${\displaystyle (c-\delta _{1} ,c+\delta _{1} )}$, we have ${\displaystyle |f(p)-f(q)|< \epsilon }$.
Now setting ${\displaystyle \delta =\delta _{1}}$, ${\displaystyle (2)}$ gives ${\displaystyle 0\leq m(c-)-m(c+)< \epsilon }$ and since ${\displaystyle \epsilon  >0}$ is arbitrary, the result follows. If $\displaystyle c$ is end point $\displaystyle a$, then set $\displaystyle m( c-) =m( a)$ and the proof still works! Similarly for the other end.
Is my proof correct? Thanks.
 A: In Case (2), how do you know that $t\in (c-\delta,x)$? It seems like $t$ is defined by $f(t)=m(x)$ so all you can say about it a priori is that $t\in [a,x]$. Similarly all you know about $z$ is that $x<z\leq y$, but you seem to want to claim that  $t,z\in (c-\delta_1,c+\delta_1)$ so that you can conclude $|f(t)-f(z)|<\epsilon$.
In Case 2, you can clean up the argument considerably by using the intermediate value theorem to conclude that $f(z')=m(x)$ for some $z'\in [x,z]$ (this follows because $f(x)>m(x)>f(z)$). By continuity of $f$, we have $|f(z')-f(z)|<\epsilon$ provided that $|x-y|<\delta_{\epsilon}$. But also $|f(z')-f(z)|=|m(x)-m(y)|$.
Also it seems like you are making things needlessly complicated by using one-sided limits, since you end up just bounding $m(x)-m(y)$ directly anyway.
A: This is just for fun.
For any $y \ge x$, we have $$ \min_{ x \le t \le y} f(t) =f(x)-\max_{ x \le t \le y} \big( f(x)-f(t) \big) \ge f(x)- \max_{x \le t \le y} |f(t)-f(x)| \ge m(x) -\max_{x \le t \le y} |f(t)-f(x)|$$
Thus, by definition
$$m(x) \ge m(y)= \min\big( m(x), \min_{x \le t \le y} f(t) \big) \ge m(x) -\max_{x \le t \le y} |f(t)-f(x)|$$
Hence
$$0 \le m(x)-m(y) \le \max_{x \le t \le y} |f(t)-f(x)| \le \max_{ u,v\in[x,y]} | f(u)-f(v)| $$
Which gives the conclusion.
