Show that if $f$ is continuous at $a$ and $f(a)≠0$ then $f$ is nonzero in an open ball around $a$. Here is the question I'm dealing with:
Let $U$ be an open set of $\mathbb{R}^{n}$, $f:U\rightarrow\mathbb{R}^{n}$ a function and $a\in U$ a given point.
Show that if $f$ is continuous at $a$ and $f(a)\neq 0$ then there exists $r> 0$ such that $B'_{r}(a)\sqsubseteq$ $U$ and $f(x)\neq0$ for each $x\in$ $B'_{r}(a)$.
I am not sure of anything I did so far but I think since $f$ is continuous and $f(a)\neq 0$, $|f(a)|>0$ for every $a$. If I choose $r$ as $1/n$ for some 
$n\in\mathbb{N}$ (archimedean property) such that $|f(a)|>1/n>0$ then $x$ would be both in $B'_{r}(a)$ and $f(x)\neq 0$. Is my way of thinking correct? I feel like I'm missing something, because I didn't use f being continuous as I should.
 A: $V=\mathbb{R}^n \setminus \{0\}$ is open. As $f$ is continuous $f^{-1}(V)$ the inverse image under $f$ is open. And also contains $a$. Hence there is an open ball $B(a,2r)$ centered on $a$ of radius $2r$ included in $f^{-1}(V)$. The closed ball $B^\prime(a,r)$ is included in $f^{-1}(V)$ so it's image under $f$ is included in $V$. Therefore $f(x) \neq 0$ for $x \in B^\prime(a,r)$ as was supposed to be demonstrated.
You can even get a slightly better result. $B^\prime(a,r)$ is compact. So its direct image $f(B^\prime(a,r))$ under the continuous map $f$ is compact. As $0 \notin f(B^\prime(a,r))$ the distance $d(f(B^\prime(a,r)),0)$ is strictly positive. Hence there exists $\alpha > 0$ such that $f(x) \ge \alpha > 0$ for $x \in B^\prime(a,r)$.
A: Take $g:U \rightarrow \mathbb R: x \mapsto \|f(x)\|$. The function $g$ is continuous (why?). Take the interval $I=(\tfrac 12 \|f(a)\|, \tfrac 32 \|f(a)\|)$. Because  $g$ is continuous $g^{-1}(I)$ is open and therefore a neighborhood of $a$ (because $a\in g^{-1}(I)$). Now you should easily prove the existence of the $r> 0$... (Note, that for each $x\in g^{-1}(I)$ you have $f(x) \neq 0$, why?)
A: $f$ has the property that if $x_n$ is a sequence that converges towards $a$, then $f(x_n)$ converges towards $f(a)$. 
You can assume (for contradiction) that there exists point arbitrarily close to $a$ that satisfy $f(x) = 0$. Using this, you can define a sequence $x_n \rightarrow a$. Now, use above property and derive a contradiction. I'll leave the details to you. 
A: $\left \| f(a)) \right \|\neq 0\Rightarrow \exists V$ open in $R^{n}$ which contains $f(a)$ and s.t $v\in V\Rightarrow  \left \| v \right \|> 0$. This is because the norm is continuous, i.e. the map $v\rightarrow \left \| v \right \|$ is continuous. But now since $f$ is continuous, there is a $U'$ open in $R^{n}$ s.t. $a\in U'$ and $f(U')\sqsubseteq V$. Now just take any ball contained in $U\cap U'$ and the result follows.
