is strong topology on a metric space the topology that is induced by metric? is strong topology on a metric space the topology that is induced by metric? Is a open set in weak topology also open in strong topology?
Yes, in a linear space $X$ that is normed, e.g. we have a metric $d$ from the norm, and the strong topology is the one induced by $d$ (so generated by the $d$-balls). The weak topology looks at all continuous linear functions from $(X,d)$ to the underlying field (often the reals). This set is called $X^\ast$, the dual of $X$.

The weak topology $\mathcal{T}_w$ is the smallest topology that makes all functions in $X^\ast$ continuous. By definition the $d$-topology is one of those topologies, so by minimality, $\mathcal{T}_w \subseteq \mathcal{T}_d$. This is only an equality of topologies in the finite dimensional case. Otherwise the weak topology is strictly smaller (i.e. weaker), hence the name.
If we consider two topologies;The 1st topology is greater or bigger or stronger or finner whatever you says if it contains all the open sets of 2nd topology. So you are correct.