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I've seen the proof of the title and roughly follow every steps of the proof. But I cannot see the intuition of this statement. Could someone show me a concrete example using $delta-epsilon$? Thanks!

Let me give you some examples that may help to form the intuition about compact sets. Although these examples will be functions on $Bbb R$, the idea for $Bbb R^n$ is not that different.

By the Heine-Borel theorem, compact sets in $Bbb R^n$ are precisely those that are closed and bounded. Let's see what happens when these $2$ conditions are not satisfied.

Closed: Strong oscillation is one reason that prevents a continuous function from being uniformly continuous. Consider the continuous function $f:(0,1]toBbb R$ defined by
$$
f(x)=sinleft(frac 1x right).
$$
This function is not uniformly continuous because the oscillation becomes infinitely strong near $x=0$. This happens because $x=0$ is on the boundary of $(0,1]$ but $f(0)$ need not be defined, hence the behaviour of $f$ can be very wild near the boundary points. On the other hand, if our domain is compact, says $D=[0,1]$ then $f(0)$ must be defined so the oscillation near $x=0$, if there is any, must "dies out" fast enough so that $lim_{yto 0}f(y)=f(0)$.

Bounded: Steepness is another reason that a function can fail to be uniformly continuous. Consider $f:[0,infty)to Bbb R$ befined by
$$
f(x)=x^2.
$$
Suppose that $a<b$. By the mean value theorem, we have $|f(a)-f(b)|=|f'(xi)||a-b|$ for some $xiin(a,b)$. We know that $f'(x)=2x$ and that our domain is not bounded, thus $f'(xi)$ is unbounded too. This implies that $f$ is not uniformly continuous. If, however, our domain is bounded, says $D=[0,M]$, then $f'(x)$ is bounded and our function must be uniformly continuous.

In general, $f$ need not be differentiable so the above reasoning doesn't apply literally. Still, I hope that gives some intuition that might help. The situation is also related to the extreme value theorem, i.e. that a continuous function cannot be unbounded on a compact set.

A concrete example can be as follows. Take e.g. $f(x) = x^2$. That is a continuous function on $mathbb{R}$, but it is not uniformly continuous because $|f(y) - f(x)| = |y^2 - x^2|$ cannot be kept smaller that a given $varepsilon>0$ for every $x,y$ whose difference $|y-x|$ is less that a given $delta_{varepsilon}$ depending only on $varepsilon$ but not on $x,y$. In other words, there is no $delta_{varepsilon}>0$ such that $|y-x| < delta_{varepsilon}$ implies $|f(y) - f(x)| = |y^2 - x^2| < delta_{varepsilon}$, because $|y^2 - x^2| = |x+y||x-y|$, and it does not matter how small $|x-y|$ is, the other factor $|x+y|$ can be made large enough (by picking $x,y$ large) so that $|y^2 - x^2| geq delta_{varepsilon}$. However, a compact set of $mathbb{R}$ such as the interval $[0,1]$ is bounded, and that makes the factor $|x+y|$ also bounded, more specifically $|x+y| leq 2$ on $[0,1]$. So now we can find a $delta_{epsilon} = frac{varepsilon}{2}$ such that $|y-x| < delta_{varepsilon}$ implies
$$|y^2 - x^2| = |x+y||x-y| < 2 delta_{varepsilon} = 2 cdot frac{varepsilon}{2} = varepsilon$$

Edit: By the time I finished writing my answer I found that BigbearZzz had already posted a more complete answer. I'll post mine anyway as a simpler version, and using $varepsilon-delta$ as requested by the OP.

If for say $epsilon=1$ there's no $delta$ that works in the entire compact set, then let $x_n$ be such that $delta=1/n$ doesn't work. Then an accumulation point of the $x_n$ is a point where $f$ is not continuous at all.

Consider $f(x)=frac{1}{x}$ which is continuous on $(0,infty)$ but not uniformly so.

Why can't we find one $deltagt 0$ that works for all $xin (0,infty)$? As $x$ gets closer to $0$, $ f$ gets steeper so that small changes in $x$ lead to larger changes in $f$.

Suppose we are given $epsilongt 0$ and $ain(0,infty)$. We want to find $deltagt 0$ such that $|x-a|ltdeltaimplies|frac{1}{a}-frac{1}{x}|ltepsilon$.

$$|frac{1}{a}-frac{1}{x}|ltepsilonimplies -epsilonltfrac{1}{a}-frac{1}{x}ltepsilonimplies -frac{1}{a}-epsilonlt -frac{1}{x}ltepsilon-frac{1}{a}impliesfrac{1}{a}+epsilongtfrac{1}{x}gtfrac{1}{a}-epsilon$$

The right hand side is equivalent to $frac{1+aepsilon}{a}gtfrac{1}{x}gtfrac{1-aepsilon}{a}$.

If $epsilon$ is large enough that $1-aepsilonlt 0$, we can restrict $epsilon$ to be smaller so that $1-aepsilongt 0$. We can do this because if the absolute value is less than the smaller $epsilon$ then it must also be less than the larger $epsilon$.

Restricting $epsilon$ and taking reciprocals gives us $frac{a}{1+aepsilon}lt xltfrac{a}{1-aepsilon}$.

For small $epsilon$, $ frac{a}{1+aepsilon}$ is a number slightly smaller than $a$ and $frac{a}{1-aepsilon}$ is a number slightly larger than $a$. Therefore, as $a$ gets closer to $0$, we require a smaller $delta$ to make $|frac{1}{a}-frac{1}{x}|ltepsilon$.