Sequence implies accumulation
Let $X$ be an arbitrary topological space.
Let $S\subseteq X$ and let $x\in S$.
Let $(s_n)$ be a sequence in $S\setminus\{x\}$ converging to $x$.
Let $U$ be an open set containing $x$. Then by the definition of convergence, there is an $N$ such that whenever $n \ge N$, $s_n \in U$. In particular, $s_N\in U$. Since $(s_n)$ is a sequence in $S\setminus \{x\}$, $s_N \in S\cap U$ and $s_N \ne x$. That is, every open neighborhood of $x$ contains an element of $S$ not equal to $x$, so $x$ is an accumulation point of $S$.
Accumulation implies sequence in a metric space:
Let $(X,d)$ be a metric space, let $S\subseteq X$, and let $x$ be an accumulation point of $S$.
For each positive integer $n$, let $B_n$ be the open ball about $x$ with radius $\frac1n$. Since $x$ is an accumulation point of $S$, $S\cap B\setminus\{x\}$ is non-empty for each $n$. Thus by the axiom of countable choice, there is a sequence $(s_n)$ in $S$ such that $s_n\in B_n$ for each $n$. Then since the set of all these open balls form a neighborhood basis at $x$, $(s_n)$ converges to $x$.
Accumulation does not imply sequence in a general topological space:
Let $S=\omega_1$ be the first uncountable ordinal. Let $X=\omega_1^+$ be considered under the order topology. Then $\omega_1$ is clearly an accumulation point of $S$. Suppose $(s_n)$ is a sequence in $S$. Then $(s_n)$ is bounded above by its union, which is a countable union of countable ordinals, and hence a countable ordinal. Thus $\bigcup_n s_n<\omega_1$, but $s_n$ never exceeds it, so $(s_n)$ does not converge to $\omega_1$.
Edit: The fact that a countable union of countable sets is countable is called the countable union condition, and is a form of the axiom of choice weaker than the axiom of countable choice but stronger than the axiom of countable choice for collections of finite sets.
As Dave L. Renfro said, the sentence "We know this because all rationals and irrationals are reals" is strange. We work here in the context of real line: there is nothing but real numbers, the real line is our Universe. Rational and irrational numbers were defined within this Universe, so saying they belong to it is redundant.
All you really need to do there is to argue that for every $x\in\mathbb R$ and every $\epsilon>0$ the interval $(x-\epsilon,x+\epsilon)$ contains an irrational number. This can be done by a cardinality argument: the interval is an uncountable set, while rational numbers are countable. There are more explicit ways too, like considering the set $\{ \frac{m}{n}\sqrt{2}:m\in\mathbb Z\}$ where $n$ is large enough so that $\sqrt{2}/n<\epsilon$. This set has to have a point in the interval.
Best Answer
I'll prove that $1+1/m$ is not an accumulation point of the set $$ S=\left\{1+\frac{1}{n}\;\middle|\;n\in\mathbb{N},n>0\right\} $$ All I need to find is $\varepsilon>0$ such that $$ \left(1+\frac{1}{m}-\varepsilon,1+\frac{1}{m}+\varepsilon\right) \cap S=\left\{1+\frac{1}{m}\right\} $$ that is, the only solution of $$ 1+\frac{1}{m}-\varepsilon<1+\frac{1}{n}<1+\frac{1}{m}+\varepsilon $$ is $n=m$.
The two inequalities are the same as $$ \frac{1}{m}-\varepsilon<\frac{1}{n}<\frac{1}{m}+\varepsilon $$ It is not restrictive to assume $\varepsilon<1/m$, so the inequalities become $$ \frac{m}{1+m\varepsilon}<n<\frac{m}{1-m\varepsilon} $$ and it suffices to show we can take $\varepsilon$ so that $$ \frac{m}{1-m\varepsilon}-\frac{m}{1+m\varepsilon}<1 $$ because in this case the interval cannot contain more than one integer (and it always contains $m$).
This becomes $$ m+m^2\varepsilon-m+m^2\varepsilon<1-m^2\varepsilon^2 $$ that is $$ m^2\varepsilon^2+2m^2\varepsilon-1<0 $$ which is certainly satisfied for $$ 0<\varepsilon<\frac{-m^2+\sqrt{m^4+m^2}}{m^2}=\sqrt{1+\frac{1}{m^2}}-1 $$
An explicit value is $\varepsilon=\frac{1}{2m^2}$.