Existence of a special kind of continuous injective function $f\colon A \to \mathbb R$, where $A$ is countable, relating to connectedness Let $A \subseteq \mathbb R$ be a countable set ($A$ induced with usual subspace topology), then does there necessarily exist a continuous injective function $f\colon A \to \mathbb R$ such that for every $a \in A$, there exist a connected subset $S\subseteq \mathbb R$ (with more than one point) such that $\{a\}=f^{-1}(S)$ ?I can prove the existence of such function if continuity is not required; if continuity is also required I am totally stuck. Please help. Thanks in advance

Okay, this is incomplete and hand wavy but:
For a dense real subset A (wolog $\mathbb Q$) I think we can prove a quasi Intermediate value Thereom: If continuous $f: \mathbb Q \rightarrow \mathbb R$. for $q, r \in A$ for any real $t$ between  $f(q)$ and $f(r)$ and any real $\epsilon > 0$, there exists a $v \in A$ between $q$ and $r$ where $|f(v) - t| < \epsilon$
Then an immediate corollary is that any continuous injective function is either increasing or decreasing.
Then ......
Let $f:\mathbb Q \rightarrow \mathbb R$ be continuous with the condition.
Let $r \in \mathbb Q$ Let $f(r) \in S_r$; a connected set such the $f^{-1}(S_r) = \{r\}$ with an element $y \in S_r; y \ne f(r)$. 
Wolog $f(r) < y$ so $[f(r), y] \subset S_r$ and this interval contains no $f(q); q \in A; q \ne r$.
Let $\epsilon = |f(r) - y|$.  As f is continuous, there is a $\delta > 0$ such that $q \in (r - \delta, r + \delta)\cap \mathbb Q$ (of which there are infinite) then $f(q) \in (f(r)-\epsilon, f(r) + \epsilon = y)$ but $f(q) \not \in [f(r), y]$ so $f(q)  \in (f(r)-\epsilon, f(r))\subset S_r$.  
That will apply for  $q_1 \in (r - \delta, r)$ and $q_2 \in (r, r + \delta); q_1, q_2 \in \mathbb Q$. 
Thus we have the contradiction: $q_1 < r < q_2$ but $f(q_1)$ and $f(q_2)$ are both less than $f(r)$ so $f$ is not increasing nor decreasing.
