Let $f : \mathbb{R}^n \rightarrow \mathbb{R}$ a $\mathcal{C}^1$ function such that there exists $\alpha > 0$: $$ \alpha \|x-y\| \leq \|f(x)-f(y)\|, \forall x,y \in \mathbb{R}^n$$ 1)Show that $f(\mathbb{R}^n)$ is a closed set.
I don't know how to approach this exercise. I tried to take a Cauchy sequence $(y_n) = f(x_n)$, then as $\mathbb R^n$ is a Banach space, I immediately have that $(x_n)$ converges, thus $f(x_n)$ converges in $f(\mathbb{R}^n)$, thus is closed. But I don't know how to show this rigorously, and I am unsure if my idea is correct.
Let $f(x_n)$ be a Cauchy sequence of $f(\mathbb{R}^n)$, for every $c>0$ there exists $N$ such that for $n,m>N$ implies that $|f(x_n)-f(x_m)|<c$ this implies that $\|x_n-x_m\|<{1\over\alpha}|f(x_n)-f(x_m)|<{c\over\alpha}$ we deduce that $x_n$ is a Cauchy sequence so converges towards $x$ since $f$ is continuous, $f(x)=\lim_nf(x_n)$ and the limit of the Cauchy sequence $f(x_n)\in f(\mathbb{R}^n)$, we deduce that $f(\mathbb{R}^n)$ is closed.