In Hirsch's book "Differential Topology," he claims in Chapter 2, Theorem 1.4 that the set of $C^1$-embeddings is open in the strong Whitney topology $C^1(M, N)$ where $M$ and $N$ are $C^1$ manifolds. My question is: How do you see that the set $\mathcal{N}\_1$ is open? See below for the setup and description of $\mathcal{N}\_1$. 


$M$ and $N$ are $C^1$ manifolds and we have an embedding $f: M \to N$. We equip $M$ with a locally finite atlas $\{\phi\_i, U\_i\}$ of $M$ which admits compact sets $K\_i \subseteq U\_i$ whose interiors cover $M$ too. We find a cover $\{\psi\_i, V\_i\}$ of $N$ so that $f(U\_i) \subseteq V\_i$. A lemma says we can find numbers $\epsilon\_i > 0$ so that any $C^1$ function $g: M \to N$ that is in the neighborhood 
$$\mathcal{N}\_0 = \{ g : M \to N \,\vert\, (\forall i) (\forall x \in K\_i)(g(K\_i) \subseteq V\_i)\text{ and }\, \lvert D^k(\psi\_i f\phi\_i^{-1} (x) - D^k(\psi\_i f \phi\_i^{-1}(x) \rvert < \epsilon\_i \text{ for } k = 0,1 \}
$$  

satisfies $g\vert\_{\mathrm{Int}(K\_i)}$ is an embedding. 


Since $f$ is an embedding, we can find open sets $A\_i$, and $B\_i$ in $N$ so that $f(K\_i)\subseteq A\_i$ and $f(M\setminus U\_i) \subseteq B\_i$ and $A\_i\cap B\_i = \emptyset$. The claim is that there is an open set $\mathcal{N}\_1$ in $C^0(M,N)$ about $f$ such that if $g \in \mathcal{N}\_1$ then $g(K\_i) \subseteq A\_i$ and $g(M\setminus U\_i) \subseteq B\_i$. 


The result is certainly true if $M$ is compact, so I believe it is possible to extend it to the case where $M$ is not compact using the strong Whitney topology. I just can't get it to work. 



The set of embeddings is open in the strong Whitney topology
A proof of this is also contained in Michor: Manifolds of mappings. There Proposition 5.3 establishes that the $C^r$-embeddings are $WO^1$ open in the space $C^r (M,N)$ (no restriction on the manifolds $M,N$). Thus if you set $r=1$ this implies that they are open with respect to the strong Whitney topology, as the $WO^1$-topology in Michor is coarser then the strong $C^1$-Whitney topology. For definitions and a comparison of the topologies see the book, it can be downloaded for free here: <https://www.mat.univie.ac.at/~michor/manifolds_of_differentiable_mappings.pdf>.


Going a bit in the details concerning the proof (answering to the comment by the OP, you have to decide whether this is an easy way to see that it is open):
Michor defines the set
 $$B = \{g \in C^1(X,Y) : d(g(x),f(x)) <\varepsilon (x), d\_1 (j\_1 f (x),j\_1 g (x))< \delta (x), g(K\_\alpha) \subseteq A\_\alpha , g(X\setminus U\_\alpha ) \subseteq B\_\alpha \text{for all }x \in X, \text{ for all } \alpha \}.$$


* For the first two conditions $d(g(x),f(x)) <\varepsilon (x), d\_1 (j\_1 f (x),j\_1 g (x))< \delta (x)$ recall that $d,d\_1$ are metrics adapted to the jet-bundles and $\varepsilon, \delta$ are continuous functions. Thus these conditions are open in $WO^1$ by the characterisation of the $WO^1$-topology in Michor 4.4.2.
* I claim that the conditions $g(K\_\alpha) \subseteq A\_\alpha , g(X\setminus U\_\alpha ) \subseteq B\_\alpha$ are open conditions in the $WO^1$-topology. To see this recall that $K\_\alpha$ is compact, the $A\_\alpha, B\_\alpha$ and $U\_\alpha$ are open with $K\_\alpha \subseteq U\_\alpha$ and the $U\_\alpha$ form a locally finite-family. Now $WO^1$ is induced by an embedding from the so called $LO$-topology (recorded as 4.6 in Michor) and for the $LO$-topology, 3.7 Lemma in Michor states that the above conditions are open.


As pointed out by the OP the second point is still missing details. Since the family of closed sets $X \setminus U\_\alpha$ will in general not be locally finite. We can not use for this condition as it is written the cited result. The additional information making this possible is that we have an embedding f to construct a family of neighborhoods which is locally finite and implements a condition similar to the one asked in Michors book. A writeup on how to do this can be found her <https://www.math.uni-hamburg.de/home/latschev/lehre/ws17/embeddings.pdf>


Thus in conclusion the set B is open in $WO^1$. I admit it is not pretty and easy, but that is the shortest explanation I could come up with in limited time. 


