Use continuity to prove inequality in $\textbf{R}^n$

Question

Suppose $f:\textbf{R}^n \rightarrow \textbf{R}^m$ is continuous at $a \in \textbf{R}^n$. And for some $b \in \textbf{R}^m$, $|b-f(a)|=d>0$. Prove that

$(1).$$\exists r>0$ s.t. $|f(x)-b| \geq d/2$ for all $x$ in the open ball $B_r(a)$.

$(2)$. If $f(a) \ne 0$,we have $f(x) \ne 0$ for all $x$ in some neighborhood of $a$.

I know that when $(1)$ is validated, the result of $(2)$ follows immediately. And the triangle inequality is needed to prove $(1)$.The difficulty for me is that I don't know how to use continuity in $\textbf{R}^n$.

Answer 1

By continuity, for each $\varepsilon>0$ there exists $\delta>0$ such that $|a-x|<\delta$ implies $|f(a)-f(x)|<\varepsilon$. If you take $r$ as the $\delta$ corresponding to $\varepsilon=d/2$, the triangle inequality will finish it for you.