This is the definition of a restriction of a vector bundle given in John Lee's Introduction to Smooth Manifolds.
Let $E,M$ be smooth manifolds with or without boundary and $\pi:E \to M$ a rank-$k$ smooth vector bundle and $S$ be an embedded submanifold with or without boundary of $M$. Then we have the restriction of $E$ to $S$ given by the set $E|_S = \bigcup_{p \in S} E_p$ with the projection $E|_S \to S$ obtained by restricting $\pi$.
If $\Phi: \pi^{-1}(U) \to U \times R^k$ is a local trivialization of $E$ over $U \subset M$ it restricts to a bijective map $\Phi|_U : (\pi|_S)^{-1}(U \cap S) \to (U \cap S) \times R^k$ and these form local trivializations for a smooth vector bundle structure on $E|_S$ by the chart lemma.
My question is if $S$ is embedded, is $E|_S$ also embedded in $E$?
I think this is true, because from the chart lemma, first for $E$, we have for each $p \in M$, choose some $U_\alpha$ containing $p$ and choose a smooth chart $(V_p,\phi_p)$ for $M$ such that $p \in V_p \subset U_\alpha$ and let $\hat{V_p} = \phi_p(V_p) \subset R^n$ or $H^n$ (where $n$ is the dimension of $M$). Define a map $\tilde{\phi_p}:\pi^{-1}(V_p)\to \hat{V_p}\times R^k$ by $\tilde{\phi_p}=(\phi_p \times Id_{R^k} )\circ \Phi_\alpha :$ $$ \pi^{-1}(V_p)\to V_p\times R^k \to \hat{V_p}\times R^k.$$ Then $\{(\pi^{-1}(V_p),\tilde{\phi_p}):p\in M\}$become the charts for $E$ and the topology for $E$ is given by taking all sets of the form $\tilde{\phi_p}^{-1}(\hat{V_p}\times W)$ where $W$ is open in $R^k$ as a basis.
Now we can do the same for $E|_S$. Since $S$ has a subspace topology of $M$, for each for each smooth chart $(W_p,\psi_p)$ for $S$ that we choose, and the maps $\tilde{\psi_p}=\Phi|_{U_\alpha}\circ (\psi_p \times Id_{R^k})$, $W_p$ is of the form $V_p \cap S$ where $V_p$ is open in $M$. Hence the basis elements of $E|_S$ is given by $\tilde{\psi_p}^{-1}(\hat{W_p}\times U)$ where $U$ is an open set in $R^k$ and will be given by $\Phi_\alpha^{-1}((V_p\cap S)\times U) = \Phi_\alpha^{-1}((V_p \times U) \cap (S \times R^k)) = \Phi_\alpha^{-1}(V_p \times U) \cap \Phi_\alpha^{-1}(S \times R^k)$. Now $\Phi_\alpha^{-1}(V_p \times U)$ will be an open set in $E$ as $V_p \times U$ is open in $M \times R^K$, and $\Phi_\alpha^{-1}(S \times R^k)$ is just $\pi^{-1}(S)= E|_S$. So this is exactly a basis element of the subspace topology of $E|_S$ with respect to $E$.
Is this argument correct?