Let $K$ be a collection of one simplices $\sigma_1,\sigma_2,\ldots$ in $\mathbb{R}^2$ where each vertex $\sigma_i$ has vertices $(0,0)$ and $(1,1/i)$. This collection $K$ of $1$-simplicies constutes a simplicial complex. Why is $|K|\subset\mathbb{R}^2$ (called a polyhedron) not metrizable?
Given the hint in the book, it suffices to show that the first countability axiom fails. Every metrizable space is first countable. First countability requires that for each point $P$ in a topological there exists a countable sequence of neighbourhoods $(N_i)_{i\in\mathbb{N}}$ such that for each neighbourhood $N$ of $P$, there exists an $i\in\mathbb{N}$ such that $N_i\subset N$. I cannot see which point violates this condition, i.e. for which point in $|K|$ there exists no such countable neighbourhood basis.
The space $X = \lvert K \rvert$ does not have the subspace topology inherited from $\mathbb R^2$, but the topology which is defined by
Let us assume to the contrary that $X$ is first countable. Then especially $p = (0,0)$ has as a neighborhood base of open $W_n$, $n \in \mathbb N$.
For each $\sigma_i$ and each $m \in \mathbb N$ let $U_i^m = \{ (x,y) \in \sigma_i \mid x < 1/m \}$. This a basis of open neigborhoods of $p$ in $\sigma_i$.
Since $W_n \cap \sigma_i$ is an open neigborhood of $p$ in $\sigma_i$, there exists $r(n,i) \in \mathbb N$ such that $U_i^{r(n,i)} \subset W_n \cap \sigma_i$. Define
$$U^* = \bigcup_{i=1}^\infty U_i^{r(i,i) + 1}$$ This is an open subset of $X$ because the $U^* \cap \sigma_i = U_i^{r(i,i) + 1}$ are open in $\sigma_i$. Clearly $p \in U^*$. Since the $W_n$ form a neighborhood base at $p$, there exists $n$ such that $W_n \subset U^*$. Thus for $i = n$ $$U_n^{r(n,n)} \subset W_n \cap \sigma_n \subset U^* \cap \sigma_n = U_n^{r(n,n) + 1} .$$ This is false, thus are assumption was wrong.
The above proof is a variant of Cantor's diagonal argument.