Proof problem explanation 
Let $f$ be a function defined at $x$. Suppose that every sequence $p_1, p_2, p_3,\dots$ in the domain of $f$ converging to $x$ has the property that $f(p_1), f(p_2), f(p_3),\dots$ converges to $f(x)$. Prove that $f$ is continuous at $x$ by contradiction.

Proof: Suppose $f$ is not continuous at $a$. If $f$ is not continuous at $a$ then $a$ cannot be an isolated point, since every function is continuous at an isolated point of its domain. If $f$ is not continuous there is some epsilon, for which no matter how what $\delta$ we choose there is a point $x_n \in S$ with $ \lVert f(x_n) − f(a)\rVert \geq \epsilon$. So let’s take
$\delta = 1/n$ and $x_n \in  S, \lVert x_n − a\rVert < 1/n$, $\lVert f(x_n) − f(a)\rVert \geq \epsilon$. Then $x_n \to a$ but $f(x_n) \to f(a)$. Hence some sequence of points $x_n$ converges to $a$ but $f(x_n)$ does not converge to $f(a)$
So a few questions for this. Can I use any arbitrary letter instead of epsilon? like $M$?
Also can anyone explain in better detail the part after If $f$ is not continuous there is some $\epsilon$. I am confused from that point on.

Your idea is good. Indeed $x$ cannot be an isolated point for the reason you state.
Suppose $x$ is an accumulation point of the domain and that $f$ is not continuous at $x$. Then there exists $\varepsilon>0$ such that, for every $\delta>0$, there exists $p$ with $\|p-x\|<\delta$ and $\|f(p)-f(x)\|\ge\varepsilon$.
(How you call $\varepsilon$ is irrelevant.)
Now we can apply the condition with $\delta=1/n$, where $n$ is a positive integer, getting a sequence $(p_n)$ with $\|p_n-x\|<1/n$ and $\|f(p_n)-f(x)\|\ge\varepsilon$.
What you need to show now is that the sequence $(p_n)$ converges to $x$, which is true because $\lim_{n\to\infty}\|p_n-x\|=0$ by squeezing.
This is the required contradiction: indeed, the hypothesis tells you that the sequence $(f(p_n))$ converges to $f(x)$ hence there exists $N$ such that, for $n>N$, $\|f(p_n)-f(x)\|<\varepsilon$ (the same $\varepsilon$ as before!).
However, by construction, $\|f(p_n)-f(x)\|\ge\varepsilon$ for all $n$.
