Can anyone help me with making the logical progression for the equivalence of these two definitions for compactness?
Topological Definition: A topological space $X$ is said to be compact if for each open covering $\{U_\alpha\}$, $\alpha \in I$, there is a finite subcovering $\{U_\beta\}$, $\beta \in J$.
(This is the definition I am used to, at least)
Now, in the book "Elementary Real and Complex Analysis" by Shilov, he gives the definition of compactness for a metric space as so:
A metric space $M$ is said to be compact if every sequence in $M$ has a limit point that is also in $M$.
I found the metric definition to be a bit odd because it is so similar to the definition of completeness.
I am assuming that these two definitions are logically equivalent, but I am not sure why. My first assumption is that Shilov is taking advantage of the Hausdorff property of the real number line, and that any compact subset of a Hausdorff space is closed. From this, by Bolzano-Weierstrass thm, we also know that any closed interval (which could be considered the closure of a neighborhood around the point $\frac{a+b}{2}$, where the interval is $[a,b]$ ) has a limit point lying within our interval. This is somewhat similar to the definition Shilov gives for locally compact:
A set is said to be locally compact if every point of $M$ has a neighborhood whose closure is compact.
Long story short: How can we deduce these definitions for metric compactness from the topological definitions? (Also we are assuming the standard Euclidean metric, or just $d(x,y) = |x-y|$ if we are in $R^1$)
Thanks!