If we look at the space $C[0,1]$ of continuous functions on $[0,1]$ we can define $\mathcal{C}:=\{\pi_{t_1,\ldots, t_d}^{-1}(B) : {t_1,\ldots, t_d} \in [0,1]$ and $B \in \mathcal{B}(\mathbb{R}^d)\}$, where $\pi_{t_1,\ldots, t_d}:C[0,1]\to \mathbb{R}^d$.
On the other hand we can tread $C[0,1]$ as a subspace of $\mathbb{R}^{[0,1]}$. On this space we can also define the finite-dimensional sets as $\mathcal{H}:=\{\tilde{\pi}_{t_1,\ldots, t_d}^{-1}(B) : {t_1,\ldots, t_d} \in [0,1]$ and $B \in \mathcal{B}(\mathbb{R}^d)\}$, where $\tilde{\pi}_{t_1,\ldots, t_d}:\mathbb{R}^{[0,1]}\to \mathbb{R}^d$. We can then define the "trace" collection as $\mathcal{H}_C=\{H\cap C[0,1]:H\in\mathcal{H}\}$.
Are the collections $\mathcal{C}$ and $\mathcal{H}_{C}$ the same, i.e. do we have $\mathcal{C}=\mathcal{H}_{C}$? I know that they both generate the same $\sigma$-algebra, but I am not sure if this equality holds.
Yes, this property is very standard (It must have a name, I just don't remember).
More generally:
($\sigma^{\vphantom{l}}_A$ means that we take a sigma-algebra as a family of subsets of $A$).
Proof: since $\sigma(\mathcal C)\cap_* A$ is a $\sigma$-algebra of subsets of $A$, we have the inclusion $\subset$ in $(1)$. Vice versa, define $$ \mathcal S = \{B\in \sigma(\mathcal C)\mid B\cap A \in \sigma^{\vphantom{l}}_A(\mathcal C_A)\}. $$ It is easy to see that this is a $\sigma$-algebra and that $\mathcal C\subset \mathcal S$. Hence we have $\supset$ in $(1)$, qed