Consider $l^{\infty}(\mathbb{R})$ the subset of $\mathbb{R}^{\mathbb{N}}$ of all bounded sequence of real numbers. Consider in $l^{\infty}(\mathbb{R})$ the restriction of the product metric, i.e, $$ d(X,Y)=\dfrac{1}{2^i}\dfrac{|x_i-y_i|}{1+|x_i-y_i|} $$
Claim 1: $(l^{\infty}(\mathbb{R}, d))$ is complete.
In fact, let $ (X_n)$ be a Cauchy sequence in $(l^{\infty}(\mathbb{R}, d)).$ So given $\epsilon>$ there is $N$ s.t $m,n>N$ implies $d(X_n,X_m)<\epsilon.$ In particular for each fixed $i$ we can choose $N$ in such way that $m,n>N$ implies $|x_n^i-x_m^i|<\epsilon.$ Therefore for each fixed $i$ the sequence $(x^i_n)_n$ is a Cauchy sequence in $\mathbb{R}$, so this sequece must have a limit $x^i$. Set $X=(x¹,x²,\ldots, x^i,\ldots)$. We claim that $X\in l^{\infty}(\mathbb{R})$, in fact observe that $$ |x^i|\leq|x^i-x_n^i|+|x_n^i| $$ and $$ |x^i-x_n^i|\leq|x_m^i-x^i|+|x_n^i-x_m^i|. $$ So we can choose $N$ s.t $m,n>N$ imples $|x_m^i-x^i|<\epsilon$, $|x_n^i-x_m^i|<\epsilon$ which implies $$ |x_i|\leq 2\epsilon+ K_n $$ for all $i$ where $K_n=\sup_i|x_i|$, finishing the proof
Claim 2: $(l^{\infty}(\mathbb{R}, d))$ is separable. This is proved in this post: Prove that a subset of a separable set is itself separable
Claim 3: Consider the sets $\Omega_k=[-k,k]^{ \mathbb{N}}$, this sets are compacts in $(l^{\infty}(\mathbb{R}, d)).$ In fact an open cover of $\Omega_k$ has the form $\{O_{\lambda}\cap l^{\infty}(\mathbb{R})\}$ where $O_\lambda $ is an open set in the product space $\mathbb{R}^{\mathbb{N}}$, now note that $\{O_{\lambda}\}$ is an open cover of $\Omega_k$ in $\mathbb{R}^{\mathbb{N}}$ which is compact, so we can extract an finite open subcover $\{O_1, \ldots,O_k\}$ of $\Omega_k$, finally observe that $\{O_{i}\cap l^{\infty}(\mathbb{R})\}_{i=1}^k$ is a subcover to $\Omega_k$ in $(l^{\infty}(\mathbb{R}, d))$ therefore $\Omega_k$ is compact in the subspace topology.
My Goals: My desires with this reasoning is to show that some of this compacts are non empty interior. Once $(l^{\infty}(\mathbb{R}, d))$ is a Polish space and $l^{\infty}(\mathbb{R})=\bigcup \Omega_k$ we can use the Baire Theorem in order to get at last one $\Omega_k$ non empty interior.
Is this reasoning correct?
