Let $f\in\mathcal{C}^1(\mathbb{R}^2,\mathbb{R})$. Prove that $f$ is not bijective ($1-1$)

Question

I found the following problem but I'm no sure if my proof is ok. Any tips, corrections or anything really is more than welcomed.

Let $f\colon \mathbb{R}^2\to \mathbb{R}$ be continuously differentiable. Show that $f$ is not bijective.

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My proof
We will prove this for functions such that $\{x\in\mathbb{R}^2\mid \partial_{x_1}f(x)\neq 0\}\neq \emptyset$ (otherwise the proof is trivial [$f(a,y)=f(b,y)$]).
Let $\bar{x}\in \mathbb{R^2}$ such that $\partial_{x_1}f(\bar x)\neq 0$. Since $f\in\mathcal{C}^1$, we have that $\partial_{x_1}f(x)\neq 0$ for all $x$ near $\bar x$. Let's denote the set of all such $x$ as $B(\bar x)$. Since all variables on this proof can be made as close as we want to $\bar x$, we will asume they all belong to $B(\bar x)$.
Let $a\in \{\alpha\in \mathbb{R}\colon 0<|\bar{x}_2-a|<\delta_\varepsilon\}$ and $|f(\bar{x}_1,\bar{x}_2)-f(\bar{x}_1,a)|=k_a<\varepsilon$ (we know that such $a$ exists for $f$ is continuous). In other words, we can make $k_a$ as close as we want to $0$ given a $a$ sufficiently close to $x_2$.
Thus we have $$f(\bar{x}_1,\bar{x}_2)=f(\bar{x}_1,a)\pm k_a$$ Or $$f(\bar{x}_1,\bar{x}_2)=f(\bar{x}_1,a)+ K_a\tag{1}$$ Where $K_a$ can be either positive or negative and can be made as small as desired given an appropriate $a$.
Let $\epsilon>0$ and let $d\in C=B[0,\epsilon]\subset \mathbb{R}$ and $f(\bar x_1+d,a)-f(\bar x_1,a)=R_d$ for the same reason, $R_d$ can be made as small as desired given a $d$ sufficiently close to $0$.
So $$R_d=f(\bar x_1+d,a)-f(\bar x_1,a)\tag{2}$$ Since $\partial_{x_1}f(\bar x_1,a)\neq 0$ (and thus $f(\bar x_1,a)$ is not a minimum or maximum with respect to $x_1$), $R_d$ ought take on both positive and negative values for some $d$. Now, choose $a$ such that $\displaystyle\min_{d\in C}R_d\leq K_a\leq \max_{d\in C}{R_d}$. Since $R_d$ is a single variable continuous function (with respect to $d$) (this also justifies the existence of $\max_{d\in C}R_d$ and $\min_{d\in C}R_d$) we have that there is a $d$ such that $R_d=K_a$ (I.V.T.). We conclude that, for this values of $a$ and $d$ (the ones such that $R_d=K_a$), $f(\bar x_1,\bar x_2)=f(x_1+d,a)$ (this follows from (1) and (2)) thus $f$ is not bijective $\square$.

Answer 1

That's a very long proof. Wlog let $f\big((0,0)\big)=0$. Take two points $x,y\in\mathbb{R}^2$ with $f(x)<0$ and $f(y)>0$ and finally let $C$ be a curve from $x$ to $y$ not containing $(0,0)$. Since $C$ is a compact connected subset of $\mathbb{R}^2$, $f(C)$ is a compact connected subset of $\mathbb{R}$, i.e. a closed interval. However, $f(C)$ contains $0$, which is impossible since $(0,0)$ was excluded from $C$ by construction. Contradiction.