A counterexample is the "Baire space" $\mathcal{N} = \mathbb{N}^{\mathbb{N}}$. This is one of the main examples of a Polish space: a separable, completely metrizable space.
One fact about this space is that all compact subsets have empty interior, that is, all compact subsets are nowhere dense. By the Baire Category Theorem, $\mathcal{N}$ is not the countable union of nowhere dense subsets, and so together with the above fact it cannot be $\sigma$-compact.
See also this question and its answer for more details:
In the aforementioned link in the OP, it is proven that $f$ has a unique fixed point, say $w$.
To show that for any $x\in X$, $f^{(n)}(x)\xrightarrow{n\rightarrow\infty}w$, we show that any subsequence of $\{f^{(n)}(x)\}$ admits a subsequence that converges to $w$.
Following the notation of the link, define the function $Q(x):=d(f(x),x)$. Since $f$ continuous, so is $Q$; moreover, unless $x$ is a fixed point of $f$, we have that $$Q(f(x))=d(f(f(x)),f(x))<d(f(x),x)=Q(x)$$
If $Q(f^{(n)}(x))=0$ for some $n_0$, then $f^{(m)}(x)=f^{n_0}(x)$ for all $m\geq m_0$ and so, $f^{(n)}(x)\xrightarrow{n\rightarrow\infty}f^{(n_0)}(x)=w$ since $f^{(n_0)}(x)=f(f^{(n_0-1)}(x))=f^{(n_0-1)}(x)$.
Suppose $x$ such that $Q(f^{(n)}(x))>0$ for all $n$. Then,
$$
\begin{align}
Q(f^{(n)}(x))<Q(f^{(n-1)}(x))<\ldots<Q(x)\quad \forall n\in\mathbb{N}\tag{0}\label{zero}
\end{align}$$
and so, $Q(f^{(n)}(x))$ converges. On the other hand, as $X$ is compact, any subsequence $\{f^{(n')}(x)\}$ of $\{f^{(n)}(x)\}$ admits a convergent subsequence $\{f^{(n_k)}(x)\}$. Say,
$$f^{(n_k)}(x)\xrightarrow{k\rightarrow\infty}y\in X$$
For any $n$, there is a unique $k$ such that $n_k\leq n<n_{k+1}$; hence
$$Q(f^{(n_{k+1})}(x))<Q(f^{(n)}(x))\leq Q(f^{(n_k)}(x))$$
and so, by the continuity of $Q$
$$\begin{align}
\lim_nQ(f^{(n)}(x))=Q(y).\tag{1}\label{one}
\end{align}
$$
By $\eqref{zero}$,
$$Q(f^{(n)}(x))>Q(y),\quad \forall n\in\mathbb{N}$$
We claim that $y$ is a fixed point. Otherwise, $Q(f(y))<Q(y)$. However, $Q(f(y))=\lim_k Q(f(f^{(n_k)}(x))\geq Q(y)$ which is a contradiction; hence $y$ is a fixed point, and by uniqueness $y=w$.
We have shown that any subsequence of $\{f^{(n)}(x)\}$ admits a subsequence that converges to the unique fixed point $w$ of $f$. From this, we conclude that in fact $f^{(n)}(x)\xrightarrow{n\rightarrow\infty}w$.
Edit: This is to address a comment from the OP:
Lemma: Suppose $(X,d)$ is a metric space, $a\in X$ and $\{a_n:n\in\mathbb{N}\}\subset X$. The sequence $a_n$ converges to $a$ iff any subsequence $a_{n'}$ of $a_n$ admits a subsequence $a_{n''}$ that converges to $a$.
Here is a short proof:
($\Longrightarrow$) Obvious.
($\Longleftarrow$) Suppose $a_n$ does not converge to $a$. Then, there is $\varepsilon>0$ such that for any $k\in\mathbb{N}$, there is $n_k\in \mathbb{N}$ such that $d(a_{n_k},a)\geq \varepsilon$. Without loss of generality, we may assume that $n_k<n_{k+1}$. Then $\{a_{n_k}:k\in\mathbb{N}\}$ is a subsequence of $\{a_n:n\in\mathbb{N}\}$, and no subsequence of $\{a_{n_k}\}$ converges to $a$ (for $d(a_{n_k},a)\geq\varepsilon$ for all $k$).
Best Answer
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.