$\newcommand{\n}{\lfloor{r/c}\rfloor}$We have \begin{equation*} s_{r,c}=\sqrt{\n c^2+(r-\n c)^2}. \end{equation*}
Indeed, by continuity and interchangeability of the coordinates, \begin{equation*} s_{r,c}^2=\sup_{a\in A_{r,c}} \sum_{i\ge1}a_i^2, \end{equation*} where \begin{equation*} A_{r,c}:=\Big\{a\in[0,\infty)^\infty\colon \sum_{i\ge1}a_i=r,\ c\ge a_1\ge a_2\ge\cdots\Big\}. \end{equation*}
For natural $i$, let
\begin{equation*}
b_i:=b_{r,c;i}:=
c\,1(i\le n)+(r-nc)\,1(i=n+1),
\end{equation*}
where
$$n:=n_{r,c}:=\n.$$
Then $b\in A_{r,c}$.
Take any $a\in A_{r,c}$. Take any real $t>0$. Then, letting $u_+:=\max(0,u)$, for some nonnegative integer $n_t=n_{a;t}$ and all natural $i$ we have
\begin{equation*}
a_i\ge t \text{ if }i\le n_t\quad\text{and}\quad a_i<t \text{ if }i>n_t.
\end{equation*}
So, if $n_t\le n$ (where $n=\n$, as above), then, letting $u_+:=\max(0,u)$, we have
\begin{equation*}
\sum_{i\ge1}(a_i-t)_+=\sum_{1\le i\le n_t}(a_i-t)
\le\sum_{1\le i\le n_t}(c-t) \\
=\sum_{1\le i\le n_t}(b_i-t)
\le\sum_{i\ge1}(b_i-t)_+
\end{equation*}
and, if $n_t>n$, then
\begin{equation*}
\sum_{i\ge1}(a_i-t)_+=\sum_{1\le i\le n_t}(a_i-t) \\
\le r-n_t t = \sum_{1\le i\le n_t}(b_i-t)
\le\sum_{i\ge1}(b_i-t)_+.
\end{equation*}
So, in any case,
\begin{equation*}
\sum_{i\ge1}(a_i-t)_+\le\sum_{i\ge1}(b_i-t)_+
\end{equation*}
for all real $t>0$.
Noting now that $u^2=2\int_0^\infty dt\,(u-t)_+$ for real $u\ge0$, we have \begin{equation*} \sum_{i\ge1}a_i^2=2\int_0^\infty dt\,\sum_{i\ge1}(a_i-t)_+ \\ \le2\int_0^\infty dt\,\sum_{i\ge1}(b_i-t)_+=\sum_{i\ge1}b_i^2. \end{equation*} So, $b$ is a maximizer of $\sum_{i\ge1}a_i^2$ over $a\in A_{r,c}$.
Thus, \begin{equation*} s_{r,c}=\sqrt{\sum_{i\ge1}b_i^2}=\sqrt{nc^2+(r-nc)^2} \\ =\sqrt{\n c^2+(r-\n c)^2}. \quad\Box \end{equation*}
In particular, $s_{r,c}\sim\sqrt{rc}\to0$ as $c\downarrow0$. So, it is not true that $s_{r,c}\to r$ as $c\downarrow0$.