Your friend is right.
From the previously solved exercise, you can show that for arbitrary $s<t$ in $(a,b)$, $\lim\limits_{u\to s+}\frac{f(u)-f(s)}{u-s}\leq\lim\limits_{v\to t-}\frac{f(t)-f(v)}{t-v}$, so $f'(s)\leq f'(t)$.
Intuitively, "$f(x)$ tends to infinity with $x$" means that as $x$ gets large, $f(x)$ gets larger than any number you pick. Because $f(x)$ is monotonic, once it gets larger than something, it stays larger than that something.
The definition is how to prove it. Say you claim that $f(x)$ goes to infinity. I am allowed to challenge you with some number $B$. To meet the challenge you have to respond with some number $A$ that guarantees if $x \gt A,$ then $f(x) \gt B$. This is a translation of the $\epsilon - \delta$ proof of a limit to "numbers near $\infty$.
For two examples, let $f(x)=\frac x{1000}$ If I challenge you with $B=1,000,000$, you can answer $A=1,000,000,000$ and demonstrate that if $x \gt 10^9, f(x) \gt 10^6$. You can even show that whatever $B$ I name, you win if $A=1000B$. Let $g(x)=10^9-\frac 1x$. If I challenge you with $B=10^6$, you can win with any $A \gt 0$. But if I am clever enough to challenge you with $B=10^{12}$ you lose. So $g(x)$ doesn't go to $\infty$ with $x$.
You might look at Jared's answer to the problem you were given.
Added: Your questioner has told you that $f(x)$ goes to infinity with $x$, given you a $y \gt f(a)$ and asked you to find an $x$ such that $f(x)=y$ and prove it unique. To prove it exists, challenge him with $B=2y$ and he will give you an $A$ such that if $z \gt A, f(z) \gt 2y$. Now the intermediate value theorem shows there is such an $x$ in $[a,A]$ and monotonicity shows it is unique.
Best Answer
Suppose $f$ is strictly convex on $(a,b)$, let $x_1<x_2<x_3<x_4<x_5$, $x_i\in(a,b)$
By strictly convex, we have $$\frac{f(x_2)-f(x_1)}{x_2-x_1}<\frac{f(x_3)-f(x_2)}{x_3-x_2}<\frac{f(x_4)-f(x_3)}{x_4-x_3}<\frac{f(x_5)-f(x_4)}{x_5-x_4}$$
Let $x_2\to x_1^+, x_4\to x_5^-$, we have $f'(x_1)<f'(x_5)$. So $f'$ strictly increasing.
The other side is true by using Mean value theorem.
For $x_1<x_2<x_3$, $x_i\in(a,b)$ since $f$ is differentiable,
$\exists c_1\in(x_1,x_2)$, s.t. $\frac{f(x_2)-f(x_1)}{x_2-x_1}=f'(c_1)$,
$\exists c_2\in(x_2,x_3)$, s.t. $\frac{f(x_3)-f(x_2)}{x_3-x_2}=f'(c_2)$,
Since $f'$ is strictly increasing, $f'(c_1)<f'(c_2)$, hence $$\frac{f(x_2)-f(x_1)}{x_2-x_1}<\frac{f(x_3)-f(x_2)}{x_3-x_2}$$