A question on metric spaces from Rudin. I feel like I'm not cut out to be a mathematician.
I am a physics major who wants to study mathematics as well, but for some reason I just get stuck on what seems like easy problems.
This one I have been working on for like days now not getting anywhere and am close to just quitting out of frustration because I just cannot see what to do.
The question is:

Let $X$ be a metric space in which every infinite subset has a limit point. For any $\varepsilon > 0$, show that $X$ can be covered by finitely many open balls of radius $\varepsilon$.

My attempt was the following:
Let $X$ be a metric space with the above property. Suppose that for all $\varepsilon > 0$, $X$ cannot be covered by finitely many balls of radius $\varepsilon$. Since $X$ infinite, $X$ can be covered by the collection of open balls $B := \{B(x,\varepsilon)\mid x\in X\}$. Look at the set $X$. since $X$ is infinite, it has a limit point $p$. 
Then I didn't know where to go from here.
Another attempt:
Let $X$ be a metric space with the above property. Suppose $X$ is infinite, then since $X\subseteq X$, $X$ has a limit point $p_0$. take the ball $B(p_0,\varepsilon)$ and subtract it from $X$. If $X\setminus B(p_0,\varepsilon)$ is finite, we are done. If the set infinite, then it clearly has a limit point $p_{1}$. Let $B_x := B(p_{x},\varepsilon)$. Take the set $X\setminus \left( B_0 \cup B_1\right)$, if this set be infinite still then it too has a limit point. If we can keep repeating this process and have finitely many balls until we have covered all of $X$ then we are done. If not, then the only way to cover $X$ is to use infinitely many balls such that $X\setminus \left(\bigcup\limits_{n}^{\infty} B_{n}\right) = \varnothing$. Since each $B_x$ corresponds to a particular limit point, we have an infinite set of limit points, call it $E$, now since $E$ in infinite, it has a limit point $p'$.
Then I got stuck again. This is where I have been for days now. I don't know what to do, people in my class talk about problem sets taking them like 10 or 15 hours to complete. I feel like I have been on this one part to one question for like forever.
Is my answer close or is it even a valid approach? And if not, could I get a hint? Anything would be greatly appreciated.
 A: Suppose $X$ cannot be covered by finitely many open balls of radius $\varepsilon$. Let us construct a sequence as follows. Let $x_1$ be any point of $X$. Now suppose we have picked the points $x_1, \dots, x_n$. Then, $\bigcup_{i=1}^n B(x_i, \varepsilon)$ does not cover $X$. So, let $x_{n+1}$ be a point of $X$ outside of $\bigcup_{i=1}^n B(x_i, \varepsilon)$. Let $S$ be the infinite set of points $x_1, x_2, x_3, \dots$. Show that $S$ has no limit point.
A: You want to show sequentially compact (every infinite sequence of points in the space has convergent subsequence) implies totally bounded (for every $\epsilon$ a finite covering by open $\epsilon$-balls).  Given those phrases there is a lot of material online with intuition and detailed proofs.
A: Fix any $\delta>0$. Let $x_1\in X$ and $x_2$ such that $d(x_2, x_1)\geqslant \delta$, and $x_3$ such that $d(x_3, x_i)\geqslant \delta$ for $i=1,2$ and we can continue this process. Show that this process will be finished after finite number of steps. Hence $X$ can be covered by finitely many open sets.
