Let $V$ be a real separable Hilbert space with norm $\Vert\cdot\Vert_V$.
Denote $\mathcal{B}(V)$ be the topology generated by the metric $d_V(x,y)=\Vert x-y \Vert$ on $V$.
Suppose that $\mathcal{B}(\mathbb{R})$ be the borel sigma algebra of $\mathbb{R}$ (smallest sigma algebra containing all open sets of $\mathbb{R}$).
Im thinking, if restricted on $R$ the metric on $V$ is simply the natural distance considered on $R$. Can we say that $\mathcal{B}(\mathbb{R})\subseteq \mathcal{B}(V)$ as a subspace topology?, that is, $$\mathcal{B}(\mathbb{R})=\{A\cap \mathbb{R}: A \in \mathcal{B}(V)\}?$$
To identify $\mathbb R$ as a subset of $V$ take a unit vector $x$ and consider the a map $f: \mathbb R \to V$ defined by $f(t)=tx$. This is a homeomorhism from $\mathbb R$ onto the one dimensional subspace $M$ generated by $x$. Hence $B$ is Borel in $\mathbb R$ iff $f(B)$ is Borel in $M$. Use the fact that $M$ is closed in $V$ to conclude that $B$ is Borel in $\mathbb R$ iff $f(B)$ is Borel in $V$. This finishes the proof.
For the question in the title the answer is YES because $f$ is a homeomorphism.