Let $(X,d)$ be a compact metric space. Prove that there exists a number $K$ such that $d(x,y)\leq K$ for each $x,y\in X$.

Let $(X,d)$ be a compact metric space. Prove that there exists a number
$K$ such that $d(x,y)\leq K$ for each $x,y\in X$.

I'm reading Intro to Topology by Mendelson.
The problem statement is in the title.
My attempt at the proof is:
Since $X$ is a compact metric space, for each $n\in\mathbb{N}$, there
exists $\{x_1^n,\dots,x_p^n\}$ such that $X\subset\bigcup\limits_{i=1}^p
B(x_i^n;\frac{1}{n})$. Let $K=\frac{2p}{n}$. Then for each $x,y\in X$,
$x\in B(x_i^n;\frac{1}{n})$ and $y\in B(x_j^n;\frac{1}{n})$ for some
$i,j=1,\dots,p$. Thus, $d(x,y)\leq\frac{2p}{n}$.
The approach I was taking is taking $K$ to be the addition of the
diameters of each open ball in the covering for $X$, that way, for any two
elements in $X$, the distance between them must be less than the overall
length of the covering. Did I say this mathematically or are there holes I
need to fill in?
Thanks for any help or feedback!