Proving a continuous function $f:X\to Y$ is uniform continuous if $X$ is compact.

Proving a continuous function $f:X\to Y$ is uniform continuous if $X$ is
compact.

I'm reading the proof of "if there's a continuous function $f:X\to Y$
where $X$ is a compact metric space and $Y$ is a metric space, then $f$ is
uniformly continuous on $X$."
The proof proceeds thus: Take any $\epsilon\in\Bbb{R}$. For any point
$p\in Y$, there exists a $\delta_p\in\Bbb{R}$ such that
$|f(x)-f(p)|<\epsilon$ for $|x-p|<\delta_p$. Take the cover $\bigcup_{p\in
X} B(p,\displaystyle{\frac{\delta_p}{2}})$. As $X$ is compact, it contains
a finite subcover $\bigcup_{i=1}^n
B(x_i,\displaystyle{\frac{\delta_{x_i}}{2}})$. Let
$m=\min\{\displaystyle{\frac{\delta_{x_1}}{2},\frac{\delta_{x_2}}{2},\dots,\frac{\delta_{x_n}}{2}}\}$.
The proof goes on to prove that $\forall x,y\in X$,
$d(f(x),f(y))<\epsilon$ for $d(x,y)<m$.
The point of the proof is that $m$ is fixed for every particular
$\epsilon$. But is $m$ really unique? There can be multiple finite
subcovers for a particular cover. Hence, $m$ is set to vary. If $m$ has an
upper bound, then this still makes sense. However, if there are are an
infinite subcovers, then it is possible $m$ has no upper bound. I'm
getting confused here.
Thanks in advance!