Assume $A$ is an uncountable set of reals, show there exists a convergent sequence $a_n$ such that $(\forall n \ a_n\in A) \land (\forall n,m \ n\neq m\implies a_n\neq a_m)$
Please check the validity of my proof:
The equistence of such a sequence implies that there is a point $a$ in $A$ such that in any neighbour-hood around $a$, there is infinitely many points within it.
Assume for a contradiction, the contrary, that is assume: for every point in $a$ there is a neighbour-hood around $a$ in which only finitely many points lie within.
Now we will develop a method to count the elements of $A$ via the above assumption, finishing our proof by contradiction. Chose any element say $v_0$, by the assumption choose a neighbourbood which contains finitely many other points (such that $v_0$ is not the smallest or largest element of this neighbourhood.). Call this set $V_0$. Define $v_{-1}:=\text{min}(V_0)$ and $v_{1}:=\text{max}(V_0)$, find a suitable neighbourhood for points $v_1$ and $v_2$ and inductively continue.
The countably many finite sets $\cdots,V_{-2},V_{-1},V_0,V_1,V_2,\cdots$ union to make another countable set $V$. It is easy to prove $A=V$. $\square$
NEW CLEANER PROOF TO CHECK:
Notice that if there exists a neighbourhood with infinitely many elements from $A$, then the desired sequence exists by Bolzano-Weierstrass.
Assume that such a neighbourhood does not exist, chose any $v\in A$, notice the amount of $a\in A$ and $a\in V_1=(v-1,v+1)$ is finite, $V_2=(v-2,v+2)/V_1$ is finite, and generally $V_n=(v-n,v+n)/V_{n-1}$ is finite. The union of countably many finite $V_n$ is countable, the obvious bijection between $A$ and $V=V_1\cup V_2\cup\cdots$ proves $A$ is countable, we have our contradiction.