Here is example $7)$, pp. $59$ of Kechris' book "Classical Descriptive Set Theory":
Let $S_\infty$ be the group of permutations of $\mathbb{N}$. With the relative topology as a subset of $\mathcal{N}=\Bbb{N}^{\Bbb{N}}$ (Baire space), it is a topological group and it is a Polish group since $S_\infty$ is a $G_\delta$ set in $\mathcal{N}$ ... Again, $S_\infty$ is not locally compact.
I want to prove the above facts. This is my attempting:
(i) $S_\infty$ is $G_\delta$: let $INJ=\{(x_n)\mid injective\}$ and $SURJ=\{(x_n)\mid surjective\}$. Then clearly $S_\infty=INJ\cap SURJ.$ Let us prove that they are $G_\delta$:
$x\in SURJ\iff\forall n\exists m\colon=x(m)$ and hence $$SURJ=\bigcap_n\bigcup_m \Bbb{N}^{n-1}\times \{n\}\times \Bbb{N}^{\Bbb{N}-n}.$$ Moreover $x\in INJ\iff \forall n,m (n\ne m\implies x_n\ne x_m)\iff \forall n,x_n\in \Bbb{N}^{\Bbb{N}}\setminus (\bigcup_{i=1}^{n-1}\pi_i(i))$ open, where $\pi_i\colon \Bbb{N}^{\Bbb{N}}\to \Bbb{N}$ is the $i$th-projection. Hence $$INJ=\bigcap_n [\Bbb{N}^\Bbb{n-1}\times (\Bbb{N}\setminus \bigcup_{i=1}^{n-1}\pi_i(i))\times \Bbb{N}^\Bbb{N}-n]$$
(ii) $S_\infty$ is a topological group: the compatible metric is $d(x,y)=\frac{1}{2^{n+1}}$, where $n=\mathrm{inf}\{m\mid x(m)\ne y(m)\}$. Then $B(x,2^{-n})=\{y\mid x(i)=y(i) \forall i\le n,n\ge m-1\}$ and the result follows.
(iii) $S_\infty$ is not locally compact: I have no successful idea.