Going through the following:
Let $U \subset \mathbb{R}^m$ be an open set and $W \subset U$ an open set with compact closure $\overline{W} \subset U$. Let $f : U \to \mathbb{R}^n$ be a $C^1$ embedding. There exists $\epsilon > 0$ such that if $g : U \to \mathbb{R}^n$ is $C^1$ and $$ \left\lVert Dg(x) - Df(x) \right\rVert < \epsilon \text{ and } |g(x) - f(x)| < \epsilon $$ for all $x \in W$ then $\left. g \right|_W$ is an embedding
I have few bits and pieces I need to put together.
Proof. By theorem 1.1. (or rather it's proof) and compactness of $\overline{W}$ there exists $\epsilon_0 > 0$ so small that if $g \in C^1(U,\mathbb{R}^n)$ for all $x \in W$ then $\left. g \right|_W$ is an immersion.
Question 1 : Theorem 1.1. is here How is compactness used exactly here? I guess this has something to do with the definition of strong topology but I can't figure where exactly.
Therefore if the lemma is false there is a sequence $g_n \in C^1(U,\mathbb{R}^n)$ such that $$ \left\lVert Dg_n(x) - Df(x) \right\rVert \to 0 $$ and $$ \left| g_n(x) - f(x) \right| \to 0 \;\;\;\;\text{$(3)$} $$ uniformely on $W$, while for each $n$ there exist distinct points $a_n,b_n \in W$ with $g_n(a_n) = g_n(b_n)$.
Question 2. I don't understand the negation of the lemma here, also the author uses a lot $\left\lVert \cdot \right\rVert$ applied to "differentials", which are tensors in general, what metric is he using? Is it the Frobenius norm?. Because theorem 1.1. applies then we can pick an open set in the strong topology around $f$. Therefore potentially the only part we can negate is whether or not $\left. g \right|_W$ is an embedding, because is definitely an immersion.
A map between two smooth manifolds is an embedding if it is a homeomorphic immersion. But I don't understand the bit "while for each $n$ there exist distinct points $a_n,b_n \in W$ with $g_n(a_n) = g_n(b_n)$". Can you clarify what does this means in terms of negating that $\left. g \right|_W$ is a homeomorphism?
By compactness of $\overline{W}$ we may assume $a_n \to a \in U$ and $b_n \to b \in U$ as $n\to\infty$. Then $f(a) = f(b)$ by $(3)$, so $a = b$.
Question 3 It's probably very similar to question 2. The fact that for every $n$ there're distincts $a_n,b_n$ such that $g_n(a_n) = g_n(b_n)$ implies we can define two sequences $\left\{ a_n \right\}, \left\{b_n \right\}$, however how is the compactness of $\overline{W}$ used in order to assess we can construct convergent sequences? I also don't understand how $f(a) = f(b)$ is implied and most importantly $a = b$. For this latter my clue is that since $f$ is an embedding then it is in particular injective, otherwise it wouldn't be invertible and hence $f(a) = f(b)$ must imply $a = b$. For the former instead I don't understand how exactly $(3)$ is used, can you clarify?
Choosing subsequences if necessary we may assume that the sequence of unit vectors $$ v_n = \frac{a_n - b_n}{\left|a_n - b_n\right|} $$ converges to a unit vector $v \in S^{m-1}$. By uniformity of Taylor expansion (with remainder in integral form) we have $$ \frac{\left|g_n(a_n) - g_n(b_n) - Dg_n(b_n)(a_n - b_n)\right|}{\left|a_n - b_n\right|} \to 0 $$ Hence $Df(b_n)v_n \to 0$. But this sequence also goes to $Df(b)v$, which therefore is $0$. This contradicts the assumption that $f$ is an immersion.
I don't understand the use of the Taylor expansion with integral reminder, but I saw a related question (this one) and I'll read through that first and maybe I'll update. However
Question 4 How does $Df(b)v = 0$ contradicts the fact that $f$ is an immersion? Does it follows from the fact that $v \in S^{m-1} \neq 0$ and hence $\text{dim Ker} \;Df(b) > 0$? (Just from basic linear algebra essentially). So $Df(b)$ is not injective, hence not an immersion.