Does a sigma-algebra always contain all open sets? This might be an easy question, but I don't really know what to use.
Let $(S,\mathcal{S},d)$ be a seperable metric space, and let furthermore $(S,\mathcal{S})$ be standard  borel 
That is there exist a injective map so $\varphi:$ $(S,\mathcal{S} )\mapsto(\mathbb{R} ,\mathcal{B} (\mathbb{R} ))$
where  $\varphi(S)\in \mathcal{B} (\mathbb{R} )$, $\varphi\in\mathcal{M}(\mathcal{S})$ and $\varphi^{-1}$ is $\mathcal{B} (\mathbb{R})-\mathcal{S}$-measurable.
Can I without further assumptions know my metric space contains all open (or closed) sets?

You cannot.
Take $S = \mathbb{R}$, $d$ the Euclidean metric, and $\mathcal{S}$ defined as follows.  Let $E \subset \mathbb{R}$ be any non-Borel set such that $E$ and $\mathbb{R} \setminus E$ both have cardinality continuum.  Then there is a bijection $\varphi : \mathbb{R} \to \mathbb{R}$ which maps $(0,1)$ onto $E$ and $\mathbb{R} \setminus (0,1)$ onto $\mathbb{R} \setminus E$.  Take $\mathcal{S} = \{ \varphi^{-1}(B) : B \in \mathcal{B}(\mathbb{R})\}$.  This is clearly a $\sigma$-algebra and $(S, \mathcal{S})$ is standard Borel, since by construction $\varphi$ is a measurable bijection from $(S, \mathcal{S})$ to $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$.  But since $E$ is not Borel, we have $(0,1) = \varphi^{-1}(E) \notin \mathcal{S}$, even though $(0,1)$ is open in $(S,d)$.
