Function continuity outside a closed subset Let $f:M \subset \mathbb{R}^p \to \mathbb{R}^q $,continuous at $a \in M $. Show that if $f(a) \notin \overline{B} (b,r) \subset \mathbb{R}^q $, then exists $ \delta > 0 $ such as $ f(x) \notin \overline{B} (b,r) $ for $ x \in M \cap B(a, \delta)$.
Demonstration:
Let $ \overline{B} (b,r) \subset \mathbb{R}^q$. By hypothesis exists $ \epsilon > 0 $ such as $ B(f(a), \epsilon ) \cap \overline{B}(b,r) = \emptyset $.
That is $ \| f(a) - b \| > \| f(a) - f(x) \| + r > \epsilon + r $. Then,
$ \| f(x) - f(a) \| < \| f(x) - f(a) \| + r < \epsilon + r < \epsilon $ if $ r > 0 $.
Therefore exists $ \delta > 0 $ such as if $ x \in B(a,\delta) \implies f(x) \in B(f(a),\epsilon) $, but $ f(x) \notin \overline{B}(b,r) $ 
 A: Here we go.
Comments:

Let $\overline{B}(b,r)⊂\Bbb R^q$. By hypothesis exists $ϵ>0$ such as $B(f(a),ϵ)∩\overline{B}(b,r)=∅$.

This is correct. And we can even say that $\epsilon < f(a) - r$.

That is $∥f(a)−b∥>∥f(a)−f(x)∥+r>ϵ+r.$

Where are you taking $x$ from? Which ball?

$∥f(x)−f(a)∥<∥f(x)−f(a)∥+r<ϵ+r<ϵ$ if $r>0.$

You just can't have $r,\epsilon > 0$ and $\epsilon +r < \epsilon$.

Therefore exists $δ>0$ such as if $x∈B(a,δ)⟹f(x)∈B(f(a),ϵ)$, but $f(x)∉\overline{B}(b,r)$.

If you claim the existence of $\delta$, or you show it explictly, or you use some strong hypothesis like continuity of some function, of properties of $\inf$ and $\sup$ of something to assure it. I'm failing to see how this would follow from the above work.
How to do it:
Again, this result is valid for metric spaces, in general. Let's do it in $\Bbb R^n$ first, and then in arbitrary metric spaces. Now, I really recommend letting go of $\Bbb R^n$: we're not adding vectors here or multiplying vector by scalars, for anything else than the triangle inequality, so we don't need $\Bbb R^n$'s vector space structure, only its metric structure.
For $\Bbb R^n$: Let $\epsilon = \|f(a)-b\| - r > 0$ (here I'm already using that $f(a) \neq \overline{B}(b,r)$). Since $f$ is continuous in $a$, for this $\epsilon$ there is a $\delta >0$ such that $\|f(x)-f(a)\| < \epsilon$ always that $x \in M \cap B(a,\delta)$ (we're taking the intersection with $M$ because it is the functions domain.. I can't speak of $f(x)$ if $x \not\in M$). Let's prove that for any $x$ in this last set, we have $\|f(x)-b\|>r$. Look: $$\begin{align*}\|f(x)-b\| &\geq |\|f(x)-f(a)\|-\|f(a)-b\||  \\ &\geq \|f(a)-b\|-\color{red}{\|f(x)-f(a)\|}\\  &> \|f(a)-b\|-\color{red}{\epsilon}   = r.\end{align*}$$
Since $\|f(x)-b\|>r$, $f(x) \not\in \overline{B}(b,r)$ and we're done.
For metric spaces, $f:M \to N$:  Let $\epsilon = {\rm d}_N(f(a),b) - r > 0$ (here I'm already using that $f(a) \neq \overline{B}_N(b,r)$). Since $f$ is continuous in $a$, for this $\epsilon$ there is a $\delta >0$ such that ${\rm d}_N(f(x),f(a)) < \epsilon$ always that $x \in B_M(a,\delta)$. Let's prove that for any $x$ in this last set, we have ${\rm d}_N(f(x),b)>r$. Look: $$\begin{align*}{\rm d}_N(f(x), b) &\geq |{\rm d}_N(f(x),f(a))-{\rm d}_N(f(a),b)|  \\ &\geq {\rm d}_N(f(a),b)-\color{red}{{\rm d}_N(f(x),f(a))}\\  &> {\rm d}_N(f(a),b))-\color{red}{\epsilon}   = r.\end{align*}$$
Since ${\rm d}_N(f(x),b)>r$, $f(x) \not\in \overline{B}(b,r)$ and we're done.
Do compare these two proofs. In fact, I was so lazy that I just copy and pasted from the first proof and swapped the $\|\cdot\|$ with $\rm d{(\cdot,\cdot)}$.
