Proof that $\vert a - b \vert \lt c \iff -c \lt a - b \lt c$ If found the following use of inequalities in a proof for the Squeeze Theorem:
$$\vert a_n - l \vert \lt \epsilon \iff - \epsilon \lt a_n - l \lt \epsilon$$
In deciding, why this is true I came up with the following proof:
Proposition: The following statements are equivalent, that is, they are either both true or both false:
$$\vert a - b \vert \lt c$$
$$-c \lt a - b \lt c$$
Proof: Suppose that $\vert a - b \vert \lt c$. Since $a-b$ can be either positive or negative,  we have
$$\vert a - b \vert = a - b \lt c \tag{1}\label{1}$$
or
$$\vert a - b \vert = -(a-b) \lt c \tag{2}\label{2}$$.
From $\eqref{2}$ it follows that $-c \lt a - b$ and combining this with $\eqref{1}$ we have
$$-c \lt a - b \lt c$$
The converse follows easily by doing those steps backwards. $\blacksquare$
Is this correct?
 A: I don’t think you are perfectly rigorous in combining the two inequalities because they assume different conditions on $a-b$ (Once positive, once negative). What should be IMHO, is something like this :
Case 1. $a-b\geq 0$. Then since $0>-c$, We have $a-b\geq0\gt-c$. Combine it with what you have done.
Similarly treat Case 2. $a-b\lt 0$.
A: Let us give labels to some statements:
[A] $|a-b|<c$
[B] $a-b<c$
[C] $-(a-b)<c$
[D] $-c<a-b<c$
Your argument can be interpreted as:
(i) [A] $\implies$ (either [B] or [C])
(ii) (both [B] and [C]) $\implies$ [D]
They are not valid to deduce [A] $\implies$ [D].
A: First of all, you consider only the implication $|a-b|<c \implies -c<a-b<c$ but not its converse $-c<a-b<c \implies |a-b|<c$.
Also, as already mentioned in @insipidintegrator's answer, you can't combine the two inequalities in your proof since they apply to two different conditions for $a-b$.
What you actually proved is "$|a-b|<c \implies (a-b<c$ if $a-b>0$ and $-c<a-b$ if $a-b<0)$" which is weaker than the original statement. Some parts are missing to complete the proof (fortunately, they are quite trivial).

A correct proof
Let $x=a-b$. Then we will prove

Let $c>0$. $|x|<c \iff -c<x<c$.

If $x\ge 0$, then $|x|=x$. Also, note that $-c<0$ so trivially, we have $x>-c$. Therefore, $$|x|<c \iff x<c \iff -c<x \text{ and } x<c \iff -c<x<c$$
Similarly, if $x<0$, then $|x|=-x$. Also, note that $c>0$ so trivially, we have $x<c$. Therefore, $$|x|<c \iff -x<c \iff -c<x \text{ and } x<c \iff -c<x<c$$
(Formally, we use the fact that if $Q$ is true and $P$ is any statement, then "$P \iff (P\text{ and } Q)$" is true).

Other proof
We have $|x|=Max(x,-x)$. Let $c>0$. Then
$$
|x|<c \iff Max(x,-x)<c \iff x<c \text{ and } -x<c 
\iff x<c\text{ and } x>-c
$$
so
$$
|x|<c \iff -c<x<c
$$

A last proof
$|x|$ is the distance between $x$ and the origin on the real line. Therefore, if $c>0$. $|x|<c$ means that the distance between $x$ and the origin is less than $c$. This means that $x\in(-c,c)$, which is equivalent to $-c<x<c$.
A: Here is my second try. Although the proofs of @Taladris are very concise and short, I wanted to try and correct my own, even though it is formulated a little bit longer. Thanks for the comments and answers.
Proposition: The following statements are equivalent, that is, they are either both true or both false:
$$\vert a - b \vert \lt c$$
$$-c \lt a - b \lt c$$
Proof: Suppose that $\vert a - b \vert \lt c$. Since $a-b$ can be either positive or negative,  we have
$$\vert a - b \vert = a - b \lt c \tag{1}\label{1}$$
or
$$\vert a - b \vert = -(a-b) \lt c \tag{2}\label{2}$$
In $\eqref{1}$ we have that $a - b \lt c$ as well as $a-b \gt 0$ and $c \gt 0$. So consequently $-c \lt  0$. Combining this we get $-c \lt 0 \lt a-b \lt c$ as required.
For $\eqref{2}$ it follows from $-(a-b) \lt c$ that $-c \lt a-b$ and, since $a-b \lt 0$, that $0 \lt c$ and consequently $-c \lt a-b \lt 0 \lt c$.
For the converse, suppose that $-c \lt a - b \lt c$. We want to show, that by this assumption, $a - b \lt c$ and $-(a-b) \lt c$, our two cases for $\vert a - b \vert \lt c$ are also true. Now note, that our assumption consists of the statements
$$a - b \lt c \tag{3}\label{3}$$
and
$$-c \lt a-b \tag{4}\label{4}$$.
So, by $\eqref{3}$ we directly obtained the first part, while from $\eqref{4}$ it follows that $-(a-b) \lt c$. $\blacksquare$
