If a unit ball is compact then why a ball of radius 5 has to be compact too? So if I use the definition of compactness that every open cover has a finite sub-cover, then as the unit ball is compact , there exists a finite subcover. But if I increase the radius of the ball, why does it still need to be compact. Intuitively speaking can't I just take very small sized and large number open sets in such a way that there is no finite sub-cover. I know that the answer to this question is no but I don't see why?
Can someone please throw some light?
 A: Strictly positive scaling is a homeomorphism on closed balls. So either none is compact or all.
A: If $B_5(x_0)$ is a closed ball with center $x_0$ and radius 5 in a Banach space, then $f(x):=5x+x_0$ maps the unit ball continuously onto $B_5(x_0)$. If the unit ball is compact, then so is $B_5(x_0)$ (as the image of a compact set under a continuous function).
A: Take your open cover of a ball of radius 5. Shrink it by a factor of 5. It's an open cover of the ball of radius 1.  Therefore it has a finite subcover.  Now reflate the subcover by a factor of 5.  Et Voila! It covers the ball of radius 5.
A: Given an open cover of the closed radius 5 ball, scale it down by a factor of $\frac15$. You obtain an open cover of the closed unit ball. By the compactness of that ball, you can pick a finite subcover. That finite subcover, scaled back up by a factor fo $5$, is a finite subcover of the original open cover of the radius 5 ball.
A: If $\mathcal U_1 = \{ U \}$ is an open cover of the unit ball, then $\mathcal U_5 = \{ 5U \}$ is an open cover of the ball of radius $5$ where $5U = \{ 5x : x \in U \}$. And vice versa in the obvious way. 
Now the result follows: any open cover of the $5$-radius ball can be mapped to a cover of the unit ball, which has a finite sub-cover; then map back that finite sub-cover to a finite sub-cover of the $5$-radius ball.
