Shorter way to prove $|\underset{x \in A}{\text{inf}}f(x)-\underset{x \in A}{\text{inf}}g(x)|\leq \underset{x \in A}{\text{sup}} |f(x)-g(x)|$ Here is a result I found in a step of a proof (not even a lemma, and I can't even find its justification):
$$|\underset{x \in A}{\text{inf}}f(x)-\underset{x \in A}{\text{inf}}g(x)|\leq \underset{x \in A}{\text{sup}} |f(x)-g(x)|$$
Here're my questions:

*

*Is my proof correct? (I've attached it below)

*Is there a more intuitive way to do it? What's the intuition behind? Can I understand the result without the concept of taking limit?

My attempt to prove it:
Denote $sup= \underset{x \in A}{\text{sup}} |f(x)-g(x)|$ (assume it's finite), and assume both $\text{inf} f(A)$ and $\text{inf} g(A)$ are real.
For all $x \in A$, $|f(x)-g(x)| \leq sup$. By the definition of infimum, we can find a sequence $(f(x_n))$, where $x_n \in A$, such that
$$\lim_{n \to \infty }f(x_n)=\underset{x \in A}{\text{inf}}f(x):= inf_f$$
So for all $x \in A$ we have:
$$|inf_f-g(x)|\leq sup$$
Similarly for $g$ we get $|inf_f-inf_g|\leq sup$.
Am I correct?
 A: As I pointed in the comments, your proof is not correct.
Assume we don't have $\inf f(A)=\inf g(A)=-\infty$ (otherwise the sentence makes no sense) and we have $sup:=\underset{x \in A}{\sup} |f(x)-g(x)|<+\infty$ (otherwise the sentence is obvious).
The intuition is to think to the graph of $g$ as being inside a tube of radius $sup$ around the graph of $f$, i.e.
$$\forall x\in A\quad f(x)-sup\le g(x)\le f(x)+sup.$$
Take the infimum (over $x\in A$) of the three terms and you are done.
A: I do have a more properties-based way to answer it.
Now to show that $|B| \leq C$, we can show that $B\leq C$ and $-B \leq C.$
So we first show that $\inf f(x) - \inf g(x) \leq \sup |f(x) - g(x)|.$
It follows by the following:
$$\begin{align}\inf f(x) - \inf g(x)&\leq \inf f(x) + \sup -g(x)\\&= \inf f(x) + \sup (f(x) - g(x) - f(x))\\&\leq \inf f(x) + \sup (f(x) - g(x)) + \sup -f(x)\\&= \inf f(x) + \sup (f(x) - g(x)) - \inf f(x)\\&= \sup (f(x) - g(x))\\&\leq \sup |f(x) - g(x)|\end{align}$$
The $-B \leq C$ case is proved if you switch $f$ and $g.$
