Does introducing $\delta$ and $\eta$ in the solution proof do anything different from my $\epsilon'$? I'm self-teaching Multivariable Calculus and new to $\epsilon-\delta$ / $\epsilon - M$ type proofs. I want to prove a theorem related to sequences.

Let, $i\mapsto \boldsymbol{a_i}$ be a sequence and $u$ be a function of $\epsilon$ such that $u(\epsilon)\to0$ as $\epsilon \to0$.
If $$(\forall \epsilon>0)(\exists M)| n>M \implies |\boldsymbol{a_n-a}|<u(\epsilon)$$
then the sequence converges.

My proof went like this: Let
$$\epsilon '=\text{max}\{u(\epsilon),\epsilon\}$$
Then $\epsilon'>0$ and as $\epsilon \to 0$, $\epsilon' \to 0$. By hypothesis, there exists $M$, such that $n>M \implies |\boldsymbol{a_n-a}|<u(\epsilon) \leq\epsilon'$. So the sequence converges.
I felt my proof was wrong because what $\epsilon'$ does in my proof just gives another name to $u(\epsilon)$, so it feels like I'm restating the assumptions... This is what my book did:
Choose $\eta >0$. Since $\lim_{t\to 0}u(t)=0$, there exists $\delta>0$ such that when $0<t\leq \delta$ we have $u(t)<\eta$. Our hypothesis guarantees that there exists $N$ such that when $n>N$, then $|\boldsymbol{a_n-a}| \leq u(\delta) = \eta$
From what I gather, our interest is to find a value that is greater than $u$. So is my proof alright? Is the line

Since $\lim_{t\to 0}u(t)=0$, there exists $\delta>0$ such that when $0<t\leq \delta$ we have $u(t)<\eta$

equivalent to my

$\epsilon '=\text{max}\{u(\epsilon),\epsilon\}$

?
 A: If I give you some positive $\epsilon$, your first attempt will show that $|a_n-a|<\epsilon'$. But you only know that $\epsilon'\ge\epsilon$, so you cannot conclude from this that $|a_n-a|<\epsilon$ and I will not be convinced that the sequence converges. For instance, suppose that $u(\epsilon) = 10\epsilon$. It is true that $u(\epsilon)\to0$ when $\epsilon\to0$. However, if I gave you $\epsilon=1$, you would only be able to show me that $|a_n-a|<10$, which doesn't suffice because I asked for a proof of $|a_n-a|<1$.
Your second attempt is better and only requires more clarity.
If I give you an $\epsilon$, the hypothesis on $u$ will give you some $\delta>0$ such that $|u(t)|<\epsilon$ provided $|t|<\delta$. Now, for such a $\delta$ there is an $M$ satisfying $|a_n-a| < u(\delta/2)$ for $n>M$. Since $\delta/2<\delta$, we can take $t=\delta/2$ and obtain $u(\delta/2)<\epsilon$. Thus $|a_n-a|<u(\delta/2)<\epsilon$, and now I'm convinced.
Why $\delta/2$ and not simply $\delta$? Because, to conclude that $u(t)<\epsilon$, the definition of limit says that you must choose $|t|<\delta$.
