Logic of the implication in $ε$-$δ$ proofs

You wrote: $$ |(3x−1)−5| = |3x−6| = |3(x−2)| = 3|x−2|<3δ < 3(ε/3) = ε. $$ First, this ought to say $$ |(3x−1)−5| = |3x−6| = |3(x−2)| = 3|x−2|<3δ = 3(ε/3) = ε. $$

When you write $$ A = B = C= D <E =F = G $$ then you're proving that if all of the above "equals" signs and the "less-than-or-equal-to" sign are true, then $A < G$ is true.

The most basic way to prove a statement of the form "If $P$ then $Q$" is to assume $P$ and prove $Q$.

In this case $P$ is "$0<|x-2|<\delta$", and $Q$ is $|(3x-1)-5|<\epsilon$. Earlier in the proof we defined $\delta = \epsilon/3$.

So we assume $P$: Assume that the statement $0<|x-2|<\delta$ is true, and set out to show that $Q$ holds. We accomplish this by the string of inequalities

$$|(3x−1)−5| = |3x−6| = |3(x−2)| = 3|x−2|<3δ,$$

then we use $\delta = \epsilon/3$ to continue

$$3\delta = 3(\epsilon/3) = \epsilon.$$

Therefore the first term is less than the last, which is precisely $Q$.

Let's just run through the proof.

We want to prove that, for any $\epsilon > 0$, there exists $\delta > 0$ such that for all $x$, $0 < |x-2| < \delta \implies |(3x-1)-5|<\epsilon$.

Hence, we take an arbitrary $\epsilon > 0$. We now want to prove the existence of a $\delta > 0$ such that the statement "for all $x$ (...)" holds.

Note that $|(3x−1)−5| = |3x−6| = |3(x−2)| = 3|x−2|$. Now we want to construct a $\delta > 0$ such that IF $|x-2| < \delta$, THEN $3|x-2| < \epsilon$. Hence, if we take $\delta = \frac{\epsilon}{3}$, we have reached our goal: $|x-2| < \delta = \frac{\epsilon}{3}$, hence $3|x-2| < \epsilon$.