Proof on showing function $f \in C^1$ on an open & convex set $U \subset \mathbb R^n$ is Lipschitz on compact subsets of $U$

The question is as follows:

Given:

(1) function $f: U \subset \mathbb R^n ==> \mathbb R$

(2) $U$ is open and convex set

(3) $f \in C^1$ in $U$

Goal: Show that $f$ is Lipschitz on any compact subset of $U$

By now, I have various ideas come to mind, but I can't connect the dots >_<

Here are my thoughts so far:

(1) Recall definitions:

(i) $f \in C^k$ means all partial derivatives up to (and including) order $k$ exist and continuous. Here $k = 1$

(ii) a set $U$ is convex if for any 2 points $x, y$ in $U$, the segment joining $x$ and $y$ is totally inside $U$

(iii) function $f$ is Lipschitz if there is a bound $M$ such that $|f(x) - f(y)| \leq M |x-y|$

(2)By a theorem in my book:

if function $f \in C^1$ on an open & convex set $U$, then for any 2 points $x$ and $y$ in $U$, there is a point $s$ lying on the segment joining $x$ and $y$ such that $f(x) - f(y) = Df(s) * (x - y)$

I firstly have a feeling that I may need this theorem in the proof, but I can't see the connection between my desired bound $M$ with $Df(s)$. So I tried to think of other ideas and ...

(3) It turns out that by all the given information, I think if I can show function $f$ is convex, then I'm done because there is a theorem which said: a convex function on a open, convex set $U$ should be Lipschitz on $U$, thus I think it must also be Lipschitz on any subset of $U$, either that subset is compact or not.

However, how can I prove function $f$ is convex, based on given information? I have a feeling that I have to use the fact that set $U$ is convex, but then I'm stuck on how to proceed further >_>

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For $x,y\in U$ there exists $s$ which lies on the line segment between $x$ and $y$ ( this line segment is contained in $U$, since it is convex), such that: $$f(x)-f(y)=\nabla f(s)\cdot (x-y)\Rightarrow\text{( Cauchy-Schwarz )}$$ $$|f(x)-f(y)|\leq|\nabla f(s)|\cdot |x-y|$$
Now, take a compact subset $S$ of $U$, and $x,y\in S$. Then take a convex compact set $S'$ such that $S\subset S'\subset U$.This is because we want $s\in S'.$
Now, since $f\in C^1$ , the first partial derivatives are continuous, and therefore, bounded in the compact set $S'$. This implies that $|\nabla f(s)|\leq M$
Concluding, for each $x,y\in S$ we have that : $$|f(x)-f(y)|\leq M\cdot |x-y|$$ which proves that $f$ is Lipschitz in $S$.