How to prove a topologic space $X$ induced by a metric is compact if and only if it's sequentially compact? A topological space $X$ is called sequentially compact if every sequence of points in $X$ has a subsequence that converges to a point in $X$. I know it's very similar to Bolzano–Weierstrass theorem but I don't really know how to argue about it.
Any good idea? Thanks for helping :)
 A: First, prove by yourself the following:
THEOREM Let $(X,d)$ be a metric space. Then $X$ is sequentially compact if and only if every infinite subset $K$ of $X$ has an accumulation point in $X$.
From this, we can work out the following:
THEOREM Let $X$ be a compact space. Then every infinite subset $K$ of $X$ has at least one limit point in $X$.
PROOF Suppose $K$ has no limit points. Then for each $x\in K$ there is a nbhd $N_x$ for which $N_x\cap K=\{x\}$. $K$ is closed, hence it is compact. Thus there exists points $x_1,\dots,x_p$ such that $K\subseteq N_{x_1}\cup\cdots\cup N_{x_p}$. It follows $K=\{x_1,\dots,x_p\}$ is finite.
Since limit points and accumulation points coincide in metric spaces, we have one direction
Now, we have to prove the other direction. You have to prove the following:
LEMMA1 Let $X$ have the property that every infinite subset of $X$ has an accumulation point. Then for each $n$ there exists a finite number of points $x_{n,1},\dots,x_{n,p}$ such that the collection of balls $B\left(x_{n,i};\dfrac 1 n\right)$ covers $X$.
[Assume this is false. Choose $x_1\in X$; and the corresponding ball $B(x_1,1/n)$. This ball cannot cover $X$, so choose $x_2$ ($1/n$ away from $x_1$) and consider $B(x_2,1/n)$. The union of the last two balls cannot cover $X$. Continue in this manner to obtain an infinite set $\{x_1,\dots\}$ with $d(x_k,x_j)\geq 1/n$ when $x\neq j$. This must have an accumulation point $x$. Take a ball $B(x;1/2n)$ and arrive to a contradiction.] 
LEMMA2 Let $X$ have the property that every infinite subset of $X$ has an accumulation point. Then for each open cover $\{O_\alpha\}$ there exists an $\epsilon >0$ such that for each $x\in X$, $B(x;\epsilon)\subseteq O_\beta$ for some element of the cover.
[Again, by contradiction. Then for each $n=1,2,\dots$ there is an open ball around some point $B(x_n:1/n)$ not contained in any element of the cover. Let $A=\{x_1,\dots\}$. Arrive to a contradiction: if $A$ is finite, some $x$ occurs infinitely often. $x$ is in some $O_\beta$, so $B(x:\delta)\subseteq O_\beta$ for some $\delta >0$. Take $n$ such that $1/n<\delta$ and $x_n=x$. Arrive to a contradiction.
If $A$ is infinite, there is some point of accumulation $x$. $x\in O_\beta$ for some $\beta$, so $B(x:\delta)\subseteq O_\beta$ for some $\delta >0$. There are infinitely many points of $A$ in $B(x;\delta /2)$. Take $n$ such that $1/n<\delta /2$ and such that $x_n\in B(x:\delta/2)$. Then $B(x_n;1/n)\subseteq B(x;\delta)\subseteq O_\beta$ which is impossible.
THEOREM Let $X$ be a metric space such that every infinite subset has at least one point of accumulation in $X$. Then $X$ is compact.
PROOF Let $\{O_\alpha\}$ be an open covering, and take $\epsilon$ the number guaranteed in the last lemma. Choose $n$ so that $1/n<\epsilon$. There is a finite set $\{x_1,\dots,x_p\}$ such that the balls $B(x_i:1/n)$ covers $X$. By the choice of $n$, each ball is contained in some $O_{\beta_i}$. The collection $\{O_{\beta_i}\}_{i=1,\dots,p}$ is the desired finite subcover.
