I wouldn't say the proof is very well written, but it is logically correct. The definition of an increasing function is
$$\hbox{if}\quad x_1<x_2\quad\hbox{then}\quad f(x_1)<f(x_2)\ .$$
As you point out, the converse of an implication is not logically equivalent to the original implication. So we cannot assume without proof that the converse is true.
However, this proof uses not the converse but the contrapositive,
$$\hbox{if}\quad f(x_1)\not<f(x_2)\quad\hbox{then}\quad x_1\not<x_2\ ,$$
which is equivalent to the definition and is therefore automatically true. Thus:
- assume $f(x_1)=f(x_2)$;
- then $f(x_1)\not<f(x_2)$;
- so $x_1\not<x_2$;
- so $x_1\ge x_2$
and so on. Alternatively, you could view it as a proof by contradiction: let $f(x_1)=f(x_2)$; suppose $x_1<x_2$; since $f$ is increasing, we have $f(x_1)<f(x_2)$; but this is not true; so $x_1\ge x_2$.
Comment. In this case you can use the definition of an increasing function to show that the converse of the definition actually is true (try it). However as far as I can see it doesn't help you with the current problem.
No, your proof is incorrect.
A counterexample
Here is a summary of your proof. Suppose $f(a_1)>a_1$. Let $a_n=f(a_{n-1})$. Then $a_n$ converges to some number $l$ when $n\to\infty$. Then $f(l)=l$.
Here is a counterexample to your proof.
Consider function $f(x)=\begin{cases}\frac14+\frac x2&\text{if }x\in[0;\frac12)\\\frac34+\frac{x-\frac12}2&\text{if }x\in[\frac12;1]\end{cases}$
Since $f(0)=\frac14>0$, let $a_1=0$ in your proof. The sequence of $a_i$'s is $0,\frac14, \frac38, \frac7{16}, \frac{15}{32},\cdots$. Check that $a_n=\frac12-\frac{1}{2^n}$ for all $n$. $l=\lim_{n\to\infty}a_n=\frac12$.
All are fine so far. However, $f(l)=\frac34\not=l$.
The mistake in the proof is, as pointed by FShrike, the following equality does not hold necessarily,
$$ \lim_{n \to \infty }f(a_{n-1}) = f(l)$$
since $f$ may not be continuous at $l$.
A correct proof
I will assume $f$ is weakly increasing only. A strictly increasing function is also weakly increasing.
Let $A_\ge=\{x\in[0;1]\mid f(x)\ge x\}\ni 0$. Since $A_\ge$ is a non-empty set bounded from above, its least upper bound is well-defined. . Let it be $\ell$.
For any $a\in A_\ge$, $\ell\ge a$. So $f(\ell)\ge f(a)\ge a$, i.e., $f(\ell)$ is also an upper bound for $A_\ge$. Hence $f(\ell)\ge\ell$.
Applying $f$ to both sides, we get $f(f(\ell))\ge f(\ell)$, i.e., $f(\ell)\in A_\ge$. Since $\ell$ is an upper bound for $A_\ge$, $\ell\ge f(\ell)$.
So $f(\ell)=\ell$.
Best Answer
You are right, b is correct. We suppose two things, and they together imply a conclusion. For the second point notice that $$ a^3>b^3 \iff a^3-b^3>0 \iff (a-b)(a^2+ab+b^2)>0$$ But we know that $(a-b)>0$ and $(a^2+ab+b^2) \geq (a+\frac{1}{2}b)^2 \geq 0$, and $a^2+ab+b^2>0$ is in fact strict because we know that the first inequality is an equality only if $b=0$ and the second is an equality only if $2a=-b \implies a=0$ which is impossible.