Inverse function theorem: Lemma 5 from Terence Tao's blog I'm trying to understand the proof of Lemma 5 from Terence Tao's blog, i.e.,
Let $X$ be an open subset of $\mathbb R^n$ and $f:X \to \mathbb R^n$ be differentiable such that $\partial f (x)$ is invertible for all $x\in X$. Fix $x_0 \in X$ and $y_0 := f(x_0)$ and let
$$
K := \{x \in X: f(x)=y_0\}.
$$

Lemma 5: Let $H$ be the connected component of $K$ that contains $x_0$. Then $H = \{x_0\}$.

Could you have a check on my attempt?
Proof: Assume the contrary that $H \neq \{x_0\}$. Then there is a path $\gamma:[0,1] \to H$ such that $\gamma(0) = x_0$ and $\gamma(1) \in H \setminus \{x_0\}$. We have $f \circ \gamma = y_0$. Then
$$
\begin{align}
\lim_{t \to 0^+} \frac{f \circ \gamma (t) - f(x_0) - \partial f (x_0)(\gamma(t)-x_0)}{|\gamma(t)-x_0|} &= -\lim_{t \to 0^+} \frac{ \partial f (x_0)(\gamma(t)-x_0)}{|\gamma(t)-x_0|} \\
&= - \partial f (x_0) \left (\lim_{t \to 0^+}\frac{\gamma(t)-x_0}{|\gamma(t)-x_0|} \right ).
\end{align}
$$
Because $\partial f (x_0)$ is invertible, it is bijective. It follows that
$$
\lim_{t \to 0^+}\frac{\gamma(t)-x_0}{|\gamma(t)-x_0|} = 0,
$$
which is a contradiction.
 A: I formulate @user85667's idea below to better understand it.

Assume the contrary that $H$ has an element other than $x_0$.

Lemma 1: Let $E, F$ be Banach spaces and $f:E \to F$ linear continuous. Then $f$ is bounded from below, i.e.,
$$
\exists c>0, \forall x \in E: |f(x)| \ge c |x|,
$$
if and only if $f$ is injective and has closed range.

Let $A := \partial f (x_0)$. Then $A$ is invertible. By Lemma 1, there is $c>0$ such that $|A (x)| \ge c|x|$ for all $x \in \mathbb R^n$.

Theorem: A connected metric space $(E, d)$ with at least $2$ points has no isolated points.

By Lemma 2, $H$ has no isolated points. So there is $(x_n)_{n\ge 1} \subset H \setminus \{x_0\}$ such that $x_n \to x_0$. Because $f$ is differentiable at $x_0$,
$$
f(x)-f(x_0) = A(x-x_0) + o (|x-x_0|) \quad \text{as} \quad x \to x_0.
$$
Notice that $x_n \in H$ and thus $f(x_n)=y_0$ for all $n \in \mathbb N$, so
$$
A(x_n-x_0) =- o (|x-x_0|) \quad \text{as} \quad n \to \infty.
$$
On the other hand,
$$
\lim_{n \to \infty} \frac{|A(x_n-x_0)|}{|x_n -x_0|} \ge c>0,
$$
which is a contradiction.
