Suppose that $M$ is a compact smooth manifold with dimension $m$, $N$ is a smooth manifold with dimension $n$; $f:M \rightarrow \mathbb{R}^{(m+n+1)}$ is a smooth embedding, $j:N \rightarrow \mathbb{R}^{(m+n+1)}$ is a smooth map.
I want to prove that: for all $\epsilon >0$, there exists a smooth embedding $g:M \rightarrow \mathbb{R}^{(m+n+1)}$, such that $\|g(p)-f(p)\|<\epsilon$ for all $p\in M$, and $g(M)$ and $j(N)$ have no intersection.
What I have tried: I managed to use partition of unity to reduce the question to charts, also I think it probably needs Whitney Approximation Theorem, but I do not know how to use it. Can someone help me?
Let $S \subset \mathbb{R}^{m+n+1}$ be an open ball centered at $0$, then consider the continuous map: $$\begin{align*} J: S\times N \rightarrow \mathbb{R}^{m+n+1} \\ (s,q) \mapsto j(q)+s \end{align*}$$ For any fixed $q \in N$, $J$ is a translation of $S$, thus a submersion at $q$. So $J$ is just a submersion of $S\times N$ and therefore transversal to $f(M)$. According to the Transversality Theorem, for almost every $s \in S$, the map $J_s(q) = j(q)+s$ is transversal to $f(M)$.
Choose such $s$ that is non-zero and small enough. Since $dim (f(M)) +dim(N) = m+n <dim(\mathbb{R}^{m+n+1})$, "$J_s$ is transversal to $f(M)$" is equivalent to say that $f(M) \cap J_s(N) = \emptyset$. Let $g(p) = f(p)-s$ be a smooth embedding from $M$ to $\mathbb{R}^{m+n+1}$, it is not difficult to see that $g(M) \cap j(N) = \emptyset$ and such $g$ satisfied all the conditions we want.