Show that $M$ is open in $\mathbb R^{n+1}$. 
Show that $$M= \{ (x_1,x_2, \dots,x_n,x_{n+1})\in\mathbb R^{n+1} : x_1^2 + x_2^2 + \dots+x_{n+1}^2 <1 \}$$ is open in $\mathbb R^{n+1}$.

I considered trying to show this using the easiest method of defining the continuous function $$f((x_1, x_2, \dots, x_{n+1})) = x_1^2 + \dots + x_{n+1}^2 $$ but then I realised that $$M=f^{-1}([0,1))$$ i.e. the image of $M$ under $f$ is $[0,1)$, which is clearly not open.
Is there any other way I can use to show that $M$ is open in $\mathbb R^{n+1}$?
 A: Sure. Let $x\in M$. Put $\delta=1-\|x\|$. If we show that the ball of radius $\delta$ around $x$ is contained in $M,$ this implies that $M$ is open. 
The key fact is that $\|z\|=(z_1^2 +\ldots+z_{n+1}^2)^{1/2}$ is a norm. In particular, it satisfies the triangle inequality. 
So, if $\|z-x\|<\delta$, then $$\|z\|=\|z-x+x\|\leq\|z-x\|+\|x\|<\delta+\|x\|=1.$$
A: Consider the complement of $M$ and take a sequence there to show that it is closed. Since we are working in a metric space, this approach works.
A: We have the following inverse image
$$
M^c=\|\cdot\|^{-1}\left([1,\infty) \right)
$$ the subset $[1,+\infty)$ is closed in $\mathbb{R}$ thus the subset $M^c$ is closed in the metric space $\mathbb{R}^{n+1}$ ($\|\cdot\|$ is a continuous function over $\mathbb{R}$),  that is $M=\left(M^c \right)^c$ is an open subset in $\mathbb{R}^{n+1}.$
A: I propose a proof in the spirit of what you have attempted. Let us define
$$g((x_1, x_2, \dots, x_{n+1})) := x_1^2 + \dots + x_{n+1}^2 -1 $$
then 
$$M=g^{-1}((-1,1))$$
Thus $M$ is the reciprocal set of the open set $\mathbb{R}$, therefore an open set of $\mathbb{R}^n$.
