Is there always a continuous function through any denumerable (or less) set of real numbers? This is a question purely stemming from curiosity. 
If $A$ is any set of points of $\mathbb{R}^2$ such that $A$ is either finite or denumerable, then does it necessarily follow that if there exists a function $f:\mathbb{R} \to \mathbb{R}$ such that $A\subseteq R$, then there is always also a function $g:\mathbb{R} \to\mathbb{R}$ such that $A \subseteq g$ and $g$ is continuous? If so, is there always an infinite number?
Put more simply, in case I've written this incorrectly, is there always a continuous function connecting any set of points in the plane if that set is either finite or denumerable?
I'm not entirely sure what topic this falls under, it seems to cross over between set theory and analysis. 
$\textbf{Edit}$: I think I've got it, now. If $A\subseteq \mathbb{R}^2$ is an either denumerable or finite set for which there exists some function $f:\mathbb{R}\to \mathbb{R}$ such that $A\subseteq f$, then does it follow that there must exist some continuous function $g:\mathbb{R}\to \mathbb{R}$ such that $A\subseteq g$?
 A: No.
Let $$A=\{\,(\tfrac1n,1)\mid n\in\mathbb N\,\}\cup\{\,(0,0)\,\}$$
Then for continuous $g\colon \mathbb R\to\mathbb R$ with $A\subseteq g$ we'd have $\lim_{n\to\infty} g(1/n)=\lim_{n\to\infty} 1\ne 0=g(0)$.
A: For a finite set of points $P_i = (x_i, y_i)$ in the plane, this is - under the mild and obvious assuption that $x_j \ne x_i$ for $i \ne j$ - correct, one can, for example, use the polynomial 
$$ g(x) = \sum_{i=1}^n y_i \cdot \prod_{j \ne i} \frac{x-x_j}{x_i - x_j} $$
which has $g(x_i) = y_i$.
For a cuontably infitine number of points, the answer is in general no: Consider the points $P_n = (\frac 1n, n)$, where $n \in \mathbf N$, which is a countably infinite set. If there were a continuous function $g\colon \mathbf R \to \mathbf R$ with $g(\frac 1n) = n$, for all $n \in \mathbf N$, continuity would give that $\lim_n g(\frac 1n)$ exists (and equals $g(0)$), but $\lim_n n$ does not exist.
A: I think such a set exists. Recall that the graph of a continuous function $f:\mathbb R\to \mathbb R$ is closed in $\mathbb R^2$. 
To find such a set $A$, we can take something that is not a graph, say the topologist sine curve and take $A$ be a dense subset away from the vertical axis.
