Is my proof for $f$ is convex iff $f'$ is monotonically increasing correct?

I am following up on my previous question. My previous attempt for the proof was wildly incorrect (my question was how that proof was exactly my old proof was incorrect) and I have now come up with a new proof.

I have to prove:

Let $f:(a, b) to R^1$ be differentiable. Prove that $f$ is convex iff $f’$ is monotonically increasing.

What I have for the proof:

($Rightarrow$) Assume $f$ is convex in $(a, b)$. Let $a<s<t<u<b$. By Exercise 23 in Chapter 4,
begin{align}tag{14.1}
frac{f(t)-f(s)}{t-s} le frac{f(u)-f(s)}{u-s} le frac{f(u)-f(t)}{u-t}
end{align}
Since $f$ is differentiable on $(a,b)$, both $ f'(s) = lim_{t to s} frac{f(t)-f(s)}{t-s}$ and $f'(t)=lim_{u to t} frac{f(u)-f(t)}{u-t}$ exist. However, applying the Order Limit Theorem on (14.1) gives
begin{align*}
lim_{t to s} frac{f(t)-f(s)}{t-s} le lim_{u to t} frac{f(u)-f(t)}{u-t} implies f'(s) le f'(t)
end{align*}
which shows that $f’$ is monotonically increasing in $(a, b)$.

($Leftarrow$) Assume $f’$ is monotonically increasing in $(a, b)$ and $a<x<y<b$. Fix $0 < lambda< 1$. By Exercise 23 in Chapter 4, we must show that
begin{equation}tag{14.0}
f(lambda x + (1- lambda)y) le lambda f(x) + (1-lambda) f(y)
end{equation}
Denote $z=lambda x+ (1-lambda)y$.Then, $z=lambda(x-y)+y$ which implies that $lambda=frac{z-y}{x-y}$. Since $lambda>0, z-y>x-y implies z>x$. Also, $1-lambda=frac{x-y-z+y}{x-y} = frac{x-z}{x-y}$. Since $lambda<1, x-z>x-y implies z < y$. Thus, $x<z<y$. Then, (14.0) can be simplified as:
begin{align*}
f(z) &le f(y) + lambda f(x) – lambda f(y) \
lambda f(z) – lambda f(x) &le f(y) – f(z) – lambda f(y) + lambda f(z) \
lambda[f(z)-f(x)] &le (1-lambda)[f(y)-f(z)]
end{align*}
Thus, since $lambda = frac{y-z}{y-x}$ and $1-lambda = frac{z-x}{y-x}$, it suffices to show that
begin{equation}tag{14.2}
frac{f(z)-f(x)}{z-x} le frac{f(y)-f(z)}{y-z}
end{equation}
Now, as we take $zto x$ on the left of (14.2) and $yto z$ on the right of (14.2), then we have $f'(x)le f'(z)$, which holds since $x<z$ and $f’$ is monotonically increasing.

Exercise 23 in Chapter 4 in Rudin:

A real-valued function $f$ defined in $(a, b)$ is said to be convex if
$$ f left( lambda x + (1- lambda) y right) leq lambda f(x) + (1-lambda) f(y)$$
whenever $a < x < b$, $a < y < b$, $0 < lambda < 1$. Prove that every convex function is continuous.

Hint: If $f$ is convex in $(a, b)$ and if $a < s < t < u < b$, show that
$$ frac{ f(t)-f(s)}{t-s} leq frac{ f(u)-f(s)}{u-s} leq frac{ f(u)-f(t)}{u-t}.$$

Can someone please read over my proof and see if there is something that I did incorrectly? Also, specifically, is my usage of the Order Limit Theorem correct and is the argument right below (14.2) correct?