How to prove the Bolzano-Weierstrass theorem in the Euclidean space. 
Theorem: Let $A$ be a bounded infinite subset of $\mathbb{R}^l$. Then
  it has a limit point.

So this is the Euclidean version of the Bolzano-Weierstrass theorem, the thing is that I was trying to prove it by induction, but it doesn't help because in the case $l=1$ we constructed a sequence of intervals, such that each iteration has infinite number of points of the set, but what is the anlogy in $\mathbb{R}^l$?, Can someone help me to prove this please?
Thanks a lot in advance :)
 A: You can construct a sequence of hypercubes instead of intervals.
For example, taking $l=2$, lets assume (without loss of generality) that $A\subseteq [0,1]^2$.
Then, split $[0,1]^2$ into four subsquares (a square is a $2D$ cube):
$$A_{11} = [0,1/2]\times [0,1/2]\\
A_{12} = [1/2, 1]\times [0,1/2]\\
A_{21} = [0,1/2]\times [1/2, 1]\\
A_{22} = [1/2, 1]\times [1/2,1]$$
You know that one of the squares has infinitely many points. Pick that square and mark it $A_1$ and split it again.
In this way, you can construct a series of squares $$A_1\supseteq A_2\supseteq A_3\supseteq\cdots$$
And you can show that the intersection of these squares contains one point $x$. And that point must be a limit point of $A$, because it lies in $A_k$ and infinitely many other points also lie in $A_k$, so infinitely many points are at most $\frac{1}{2^k}$ away from $x$.
A: The closure of $A$ is compact. In compact metric spaces each sequence has a convergent subsequence. If you may use that result you are done.
Elaboration: Since $A$ is infinite there is a injective sequence $(a_n)_{n\in\mathbb N}\subseteq A$. Since $\overline{A}$ is compact $(a_n)_{n\in\mathbb N}$ has a convergent subsequence with limit in $\overline{A}$. Since the limit is the limit of an injective sequence of elements of $A$ it is a limit point of $A$.
A: Proving it for $l=1$ is enough to get the rest.
Let $X_n = (x_{1,n},\dots,x_{l,n})$ be an injective infinite sequence of points in the set $A$.
$x_{1,n}$ is bounded
So there is a subsequence s.t $x_{1,\phi_1(n)}$ converges
Now consider $X_{\phi_1(n)} = (x_{1,\phi_1(n)},\dots,x_{l,\phi_1(n)})$
$x_{2,\phi_1(n)}$ is bounded
So there is a subsequence s.t $x_{2,\phi_1\circ\phi_2(n)}$ converges
(Note that $x_{1,\phi_1\circ\phi_2(n)}$ also converges)
We repeat that until we find a subsequence that makes all the coordinates converge.
$(x_{1,\phi_1\circ\phi_2\dots\circ\phi_l(n)},\dots,x_{l,\phi_1\circ\phi_2\dots\circ\phi_l(n)})$ all converge, so $X_{\phi_1\circ\phi_2\dots\circ\phi_l(n)}$ also converges and its limit is a limit point of $A$
