If $f$ is a real-valued function and is convex on an open interval $(a,b)$, show that $f$ is bounded on any closed subset $[c,d] \subset (a,b)$ I read on a post that to prove a real function that is convex on an open interval is bounded on any closed subset is straightforward, so I wanted to give it a shot. However, I noticed that my approach seemed a little more convoluted than I had anticipated, which makes me think that my strategy is misguided. Any tips would be appreciated. Cheers.

Without assuming differentiability or continuity, prove the following:

If $f$ is a real-valued function and is convex on an open interval $(a,b)$, show that $f$ is bounded on any closed subset $[c,d] \subset (a,b)$

I will show that on an arbitrary $[c,d]$, $f$ has a lower-bound. The argument for an upper bound works similarly.

From the definition of continuity, we have the following:
$$\forall x_1,x_2,x_3 \in (a,b): x_1 \lt x_2 \lt x_3 \rightarrow\frac{f(x_3)-f(x_1)}{x_3-x_1} \gt  \frac{f(x_2)-f(x_1)}{x_2-x_1}$$
Rearranging the inequality, noting that $x_3 -x_1 \gt 0$, we have that:
$$f(x_3)\gt \frac{x_3-x_1}{x_2-x_1}\cdot\big[f(x_2)-f(x_1)\big]+f(x_1) \quad (\dagger_1)$$
In order to get this inequality in the desired form, we need to get rid of the $x_3$ symbol in the right-hand side of the inequality.
There are two cases: $f(x_2)-f(x_1) \geq 0$ or $f(x_2)-f(x_1) \lt 0$. Suppose the former.
Case 1: $f(x_2)-f(x_1) \geq 0$
Consider an arbitrary closed interval $[c,d]$ such that $[c,d] \subset (a,b)$.
Referencing $(\dagger_1)$, let $x_3 =x \in [c,d]$ and let $x_1, x_2 \in(a,c)$ such that $x_1 \lt x_2$. Next, consider some $k \in (a,c)$.
Firstly, note that because $x \gt k$, we have that $\frac{x-x_1}{x_2-x_1} \gt\frac{k-x_1}{x_2-x_1}$. By assumption, $f(x_2)-f(x_1) \geq 0$; therefore, we have the following:
\begin{align}f(x) \gt \frac{x-x_1}{x_2-x_1}\cdot\big[f(x_2)-f(x_1)\big] +f(x_1)\geq \frac{k-x_1}{x_2-x_1}\cdot\big[f(x_2)-f(x_1)\big] +f(x_1) \end{align}
which means that $f(x) \gt \frac{k-x_1}{x_2-x_1}\cdot\big[f(x_2)-f(x_1)\big] +f(x_1)=m$. Noting that $x$ was an arbitrary element of $[c,d]$, let $m$ be a lower bound of $f$ on $[c,d]$.
Now suppose that $f(x_2)-f(x_1) \lt 0$.
Case 2: $f(x_2)-f(x_1) \lt 0$
The set up is similar, but this time we stipulate that $k \in (d,b)$. As such, we must have that $x \lt k$. This means that $\frac{x-x_1}{x_2-x_1} \lt \frac{k-x_1}{x_2-x_1}$. By assumption, $f(x_2)-f(x_1) \lt 0$; therefore, we have the following:
\begin{align}f(x) \gt \frac{x-x_1}{x_2-x_1}\cdot\big[f(x_2)-f(x_1)\big] +f(x_1)\gt \frac{k-x_1}{x_2-x_1}\cdot\big[f(x_2)-f(x_1)\big] +f(x_1) \end{align}
which means that $f(x) \gt \frac{k-x_1}{x_2-x_1}\cdot\big[f(x_2)-f(x_1)\big] +f(x_1)=m'$.  Noting that $x$ was an arbitrary element of $[c,d]$, let $m'$ be a lower bound of $f$ on $[c,d]. \quad \square$
 A: Hint: If the function $f$ is convex, then the graph of it on $[c,d]$ is bounded from above by the straight line connecting $(c, f(c))$ and $(d, f(d))$.
For the bound "below", construct two straight lines: $l_1(x)$ connecting $(a, f(a))$ with $(c, f(c))$ and $l_2(x)$, connecting $(b, f(b))$ with $(d, f(d))$. The piecewise-linear function that goes from $(c, f(c))$ alongside $l_1$ until the intersection of $l_1$ with $l_2$ and then goes alongside $l_2$ until $(d, f(d))$ bounds $f$ from below.

(In fact, $l_1$ and $l_2$ are also bounded on $[c,d]$, being linear, so any of those two functions is a lower bound for $f$. In the above, the piecewise-linear function is actually $\max(l_1(x), l_2(x))$ and makes for a slightly finer lower bound.)
A: A much simpler argument is to use the fact that a convex function in a domain is continuous in the interior of its domain, so $f$ will be actually continuous in $(a,b)$, hence bounded on any compact subset.
A: Let $e=(a+c)/2$ and $e'=(d+b)/2$ and $e''=(e+c)/2.$
(1). Let $g=\max \{sf(e)+(1-s)f(e'):s\in [0,1]\}.$ For each $x\in [c,d]$ there exists $r\in (0,1)$ with $x=re+(1-r)e',$ so $$f(x)=f(re+(1-r)e')\le rf(e)+(1-r)f(e')\le g.$$ So $f$ on $[c,d]$ is bounded above by the number $g.$
(2).  For any $x\in [c,d]$ we have $e''=re+(1-r)x$ where $r=\frac {x-e''}{x-e}\in [\frac 1 2 , \frac {d-e''}{d-e}]\subset [0,1].$  So we have  $$\bullet\quad f(e'')\le rf(e)+(1-r)f(x).$$ Now $(1-r)^{-1}=\frac {x-e}{e-e''}\ge \frac {c-e}{e-e''}=2.$ And $f(e'')-rf(e)\ge -|f(e'')|-r|f(e)|\ge -|f(e'')|-|f(e)|.$ Therefore  by $\bullet$ we have $$f(x)\ge (1-r)^{-1}(f(e'')-rf(e))\ge 2(-|f(e'')|-|f(e)|).$$
A: You can show that f is Lioschitz on$ [c,d]$. It's done here!
Proving that a convex function is Lipschitz
