How to prove that Hilbert Cube has the Fixed Point Property without using Brouwer Fixed point theorem? So these two statements might be equivalent, but still there is supposed an easier way to prove the former without knowledge in algebraic topology
It's an exercise on my textbook after the chapter concerning compact metric spaces.
The following Lemma might help:
Let $(X,d)$ be a compact metric space, $f:X \mapsto X$ be a continous mapping, if for any $\epsilon>0$ there exists $x_{\epsilon}$ such that $d(x_{\epsilon},f(x_{\epsilon}))<\epsilon$, then $f$ has a fixed point. The proof is quite easy.
So I already know that hilbert cube is a compact metric space according to the Tychonoff theorem and the fact that the countable product of any metrizable spaces is metrizable.
Note: The textbook skipped the proof of Tychonoff theorem.
Hilbert Cube: $\prod_{n=1}^\infty[0,\frac{1}{n}]$
 A: After some thinking I'm not convinced there is a proof of this result which avoids Brouwer (or a repackaged version of Brouwer). Note that Brouwer's theorem can be embedded into this problem in the following manner: for $m\geq 1$ we have the homeomorphisms
$$B^m\cong [0,1]^m\cong \prod_{n=1}^m[0,1/n]=: C_m$$
Note that $C_m$ can naturally be identified with the subspace of $H$ consisting of sequences supported on the first $m$ entries. In fact there is a natural inclusion $\iota: C_m\to H$ given by
$$\iota(a_1,\dotsc, a_m) = (a_1,\dotsc, a_m,0,0,\dotsc).$$
Endowing $H$ with the $\ell_2$ metric it is easy to show this map is continuous. Similarly one can define a continuous projection $\pi: H\to C_m$ given by truncation at the first $m$ coordinates.
Now, $H$ having the fixed point property (FPP) means that every continuous map $f: H\to H$ has a fixed point. If $g: C_m\to C_m$ is a continuous map of hypercubes, then $\iota\circ g\circ\pi: H\to H$ is continuous and by the FPP there is a sequence ${\bf x}\in H$ such such that $(\iota\circ g\circ \pi)({\bf x}) = {\bf x}$. But then $(x_1,\dotsc, x_m) = \pi({\bf x})$ is a fixed point of $g$, and Brouwer's theorem is proven.
Conversely, we can use Brouwer's theorem to prove $H$ has the FPP. Fix a continuous map $f: H\to H$. Then for each $m$ there is a "truncated" function $f_m: C_m\to C_m$ given by $f_m = \pi\circ f\circ \iota$. By Brouwer's theorem these truncated maps have fixed points in $C_m$; let ${\bf x}^{(m)}$ be the identification of these points in $H$ and note that ${\bf x}^{(m)}$ and $f({\bf x}^{(m)})=:{\bf y}^{(m)}$ agree in the first $m$ entries by construction. Then
$$d({\bf x}^{(m)}, f({\bf x}^{(m)})) = \left(\sum_{n=1}^\infty |x_n^{(m)}-y_n^{(m)}|^2\right)^{1/2} = \left(\sum_{n=m+1}^\infty |x_n^{(m)}-y_n^{(m)}|^2\right)^{1/2}\leq \left(\sum_{n=m+1}^\infty \frac{4}{n^2}\right)^{1/2},$$
where we have used that $|x_n^{(m)}-y_n^{(m)}|\leq 1/n+1/n=2/n$. Since the series $\sum n^{-2}$ is finite the tails of the series must go to $0$, meaning that given $\epsilon>0$ there is $m$ suitably large for which $d({\bf x}^{(m)}, f({\bf x}^{(m)}))<\epsilon$. Your lemma then completes the proof.
Now, the upshot: since $H$ having the FPP implies Brouwer's theorem, the argument you desire probably does not exist. Proofs of Brouwer using only "elementary" topology are highly sought after for pedagogical purposes, but the only ones I've seen use Sperner's lemma or Simplicial Approximation Theorem, neither of which I suspect was the intent of this problem. If there is a proof that $H$ has the FPP using only "basic" facts about metric spaces (by which I mean, without using algebraic or combinatorial topology), it's yet to be discovered.
