I have tried to prove that $f(x)=(x+1)^2$ is continuous as $x$ approaches $p$.
However, I have a gut feel that what I have done is not sufficient, or that something is wrong. It might be several things that I have done that is incorrect, however, my issue is using $\delta=min(1,\epsilon/(see, below))$
Let $A\subseteq\mathbb{R}$ and let $f:A\to\mathbb{R}$. Denote $p\in A$.
Then $f(x)$ is continuous at $p$ iff for every $\varepsilon>0$, $\exists$ $\delta>0$ such that
$$0<|x-p|<\delta\implies |f(x)-f(p)|<\varepsilon.$$
Here is my attempt:
$$\left| f(x)-f(p)\right| =| (x+1)^2-(p+1)^2|=|x-p||x-p+2(p+1)|$$
(point 1)
Now if $$|x-p|<1$$
then $$|x-p||x-p+2(p+1)|<|x-p|(2|p|+3)$$
now if $$|x-p|<\epsilon/(2|p|+3)$$
then
$\left| f(x)-f(p)\right|<|x-p|(2|p|+3)<\epsilon$
therefor take $\delta=min(1,\epsilon/(2|p|+3)$
Now is this right or wrong. And what would happen if I have $|x-p|>=1$, then I would not be able to continue as I did, would that min that I have only considered the case where $|x-p|<1$ and not the case where the absolute difference between $x$ and $p$ is larger or equal to $1$ see point 1
What you have done is correct ! It is not neccesary to consider the case $|x-p| \ge 1$, since for the continuity of $f$ in $p$ you only have to consider arguments $x$ , which are "near" $p$.