Originally Posted by

**Pinkk** Hmm, thanks, I think that does help. So here is my attempt.

Given $\displaystyle \epsilon > 0$ and $\displaystyle a\in \mathbb{R}^{n}$, let $\displaystyle U=B(\epsilon , f(a))$, which is clearly an open set $\displaystyle \{y : |y - f(a)| < \epsilon \}$. Since $\displaystyle f(a)\in U$, we have that $\displaystyle a\in S = \{x : f(x)\in U\}$. So $\displaystyle a$ is an interior point of $\displaystyle S$, which means there exists $\displaystyle \delta > 0$ such that $\displaystyle |x - a|< \delta $ implies $\displaystyle x\in S$, which in turn implies $\displaystyle f(x)\in U$. Since $\displaystyle f(x)\in U$ implies $\displaystyle |f(x)-f(a)|<\epsilon$, we have that given $\displaystyle \epsilon > 0$, there exists $\displaystyle \delta > 0$ such that $\displaystyle |x - a| < \delta $ implies $\displaystyle |f(x) - f(a)| < \epsilon$, which means $\displaystyle f$ is continuous at $\displaystyle a$. Since $\displaystyle a$ was arbitrary, we have that $\displaystyle f$ is continuous on $\displaystyle \mathbb{R}^{n}$. Q.E.D.