If $c \in \overline{C}$, $p \in C^\circ$ and $t \in [0,1)$ then
the point $p(t) = p + t (c-p)$ is in $C^\circ$.
Since $p \in C^\circ$, there is some open neighbourhood $U$ of $0$ such that
$U+\{p\} \subset C$.
Now I claim that if $x \in (1-t)U+\{p(t)\}$, then $x \in C$. In
particular, $p(t) \in C^\circ$.
Let $y = {1 \over 1-t} (x-tc)$ and note that $x = (1-t)y + t c$.
Then $y-p = {1 \over 1-t}(x-tc+tp - p) = {1 \over 1-t}(x-p(t)) \in {1 \over 1-t} (1-t)U = U$.
Hence $y \in C$ and so $x \in C$.
Correction: The above is incomplete. It does not use the fact that $c \in \overline{C}$ anywhere.
Suppose $U$ is a convex neighbourhood of $0$ such that $U+\{p\} \subset C$.
Note that $c \in C+\epsilon U$ for any $\epsilon>0$. Pick some $t \in [0,1)$ and choose $\epsilon>0$ such that $\epsilon {1+t \over 1-t} \le 1$.
Now suppose $u \in U$ (so that $\epsilon u \in \epsilon U$), then
\begin{eqnarray}
p(t) + \epsilon u &=& (1-t)p+t c + \epsilon u \\
&\in& (1-t)\{p\} + t (C+\epsilon U) + \epsilon U \\
&\in& (1-t) (\{p\} + \epsilon {1+t \over 1-t} U) + t C \\
&\subset & (1-t)C + t C \\
&=& C
\end{eqnarray}
In particular, $p(t) \in p(t)+\epsilon U$, so $p(t) \in C^\circ$.
It follows immediately that $c$ is in the closure of the interior.
I have found a reference myself:
Richard B. Holmes, Geometric Functional Analysis and its Applications, (GTM 24). Page 9 and Page 59.
In the book, a related notion $\mathrm{lin}(A)$ is defined, which is the union of $A$ and points linearly accessible from $A$. Here a point $x$ is linearly accessible from $A$ if there is some $a\in A$ such that $[a,x)\subset A$. It is easy to note that for any $A$ convex, the algebraic closure defined in the question coincides with $\mathrm{lin}(A)$.
It is commented in the book on Page 9 part D that $\mathrm{lin}(A)$ is the (topological) closure of $A$ in a finite-dimensional space if $A$ is convex. A more general statement is part (c) of the Lemma on Page 59, which states
$\mathrm{lin}(A)=\bar{A}$ for any convex set $A$ with nonempty interior in a topological vector space.
If $A$ lies in a finite-dimensional space, then the statement above applies to the affine hull of $A$ (i.e. the smallest affine subspace containing $A$), implying statement (1) in the question.
Best Answer
If $M$ is any dense proper subspace of a normed linear space then $Int (M)$ is empty and $int(\overline M)$ is the whole space.
Note that such a subspace $M$ does not exist in a finite dimensional case.