Are these sets representing the subspace topology equal? I have a question regarding the representation of the subspace topology.
Let $(\Omega, \mathcal T)$ be a topological space, $\Omega' \subseteq \Omega$ and $\mathcal T' \subseteq \mathcal T$ the subspace topology with respect to $\mathcal T$. Then, by definiton $$\mathcal T' = \{U\cap\Omega' \,|\, U\in \mathcal T\}.$$
My question: Is the following representation equal to the first one? $$\mathcal T'= \{U\in \mathcal T \,|\,U\subseteq \Omega'\}$$

The second definition is not correct in general.
The subspace topology $\Omega '$ is composed by the intersection of the open sets of $\Omega$ with $\Omega'$, hence is possible that there are $U\in\mathcal T$ that are not subsets of $\Omega'$ and that at the same time we need them to define the subtopology (the open sets) of $\Omega'$.
By example take the topological subspace of $\Bbb R$ defined by $A:=[0,1)\cup(2,3]$. Then from the open sets $(-1,1)$ and $(2,4)$ of (the standard topology of) $\Bbb R$ we have that
$$A\cap (-1,1)=[0,1),\quad A\cap (2,4)=(2,3]\tag{1}$$
Hence $[0,1)$ and $(2,3]$ are open sets in the topological subspace $A$ (by your first definition of topological subspace).
But if we follow your second "definition" of topological subspace we cannot define the open sets in $(1)$ because is clear that $[0,1)$ and $(2,3]$ are not open sets in (the standard topology of) $\Bbb R$.
