Let S be any compact and denumerably infinite metric space and let d be the metric of S. We shall say that S satisfies condition C, if there exists at least one infinite sequencep(1),p(2),.....,p(n),....of points of S such that (1) each point of S occurs once and only once in the sequence (2) the series d(p(1),p(2))+d(p(2),p(3))+.....+d(p(n),p(n+1))+....is convergent. If S satisfies condition C, an infinite version of the travelling salesman problem can be posed for the points of S. S always contains at least one limit point, since otherwise it could be covered by an infinite set of pairwise disjoint open balls, each containing a point of San impossibility if S is compact. It is easy to show that S cannot satisfy condition C if it contains more than one limit point. My question is: If S contains one and only one limit point, does S then necessarily always satisfy condition C?
Let the space consist of a point $P$ and infinitely many other points $Q_n$ for $n\in\mathbb N$. Let the distance from $P$ to $Q_n$ be $1/n$. Let the distance from $Q_n$ to $Q_m$ for $m\neq n$ be $(1/n)+(1/m)$. This seems to be a counterexample for your question. 


The countable compact subspace $\ S:=\{q_0\ q_1\ \ldots\}\subseteq\ell^2\ $ of Hilbert space $\ \ell^2,\ $ given below, fails to have property C, i.e. it cannot be travelled along any finite length path (condition C was defined in the Question above):
Now let $\ (p_1\ p_2\ \ldots)\ $ be a sequence of points of $\ S,\ $ in which each point $\ q_k\ $appears exactly one time. We will see that this routing path has infinite length: $$\sum_{k=2}^\infty p_kp_{k1}\ =\ \infty$$ PROOF For each natural $\ M\ $ there exists natural $\ n\ $ such that $$\{p_1\ \ldots\ p_n\}\ \supseteq\ \{q_1\ \ldots\ q_M\}$$ Let's use the following inequality: $$\forall_{a\ b\in\ell^2}\quad \left(\ a\cdot b = 0\quad\Rightarrow\quad ab\ \ge\ \frac{a}{\surd 2}\ +\ \frac{b}{\surd 2}\ \right)$$ Thus: $$\sum_{k=2}^n p_kp_{k1}\ \ge\ \frac 1{\surd 2}\cdot\sum_{k=2}^n \left(p_{k1}+p_k\right)\ =\ \surd 2\cdot\sum_{k=1}^n p_k\ \ \frac 1{\surd 2}\cdot\left(p_1+p_n\right)$$ hence $$\sum_{k=2}^n p_kp_{k1}\ \ge\ \surd 2\cdot\sum_{k=1}^M q_k\ \ \frac 1{\surd 2}\cdot\left(\frac 11+\frac 12\right) $$ and finally: $$\sum_{k=2}^n p_kp_{k1}\ \ \ge\ \ \surd 2\cdot\log(M+1)\ \ \frac{3\cdot\surd 2}4\quad\longrightarrow\quad\infty$$ for $\ M\rightarrow\infty\ $ (where $\ n\ $ depends on $\ M$) END of PROOF



I'll show a counterexample in Euclidean $\ (\mathbb R^2\ \,d)$:
Here we understand that $\ (p(1)\ p(2)\ \ldots)\ $ being a route of $\ X\ $ means that all points $\ p(t)\ $ are different, and $$ \{p(1)\ p(2)\ \ldots\}\ =\ X$$ and the length of such route is $\ \sum_{k=1}^\infty d(p(k)\ p(k+1))$. PROOF Let $\ O:=(0\ 0)\in\mathbb R^2.\ $ Let variables $\ k\ n\ $ run only through natural numbers $\ 1\ 2\ \ldots\ .\ $ Define $$ X\ :=\ \{O\}\ \cup\ \left\{(\frac 1k\ \frac 1n)\ :\ \max(k\ n)\le 2\cdot\min(k\ n)\right\} $$ Consider $\ \ \delta(x) := \min_{y\in X\setminus\{x\}} d(x\ y)\ \ $ for every $\ x\in\mathbb R^2\setminus\{O\}.\ $ This minimum is well defined and satisfies: $$ \delta\left(\left(\frac 1k \frac 1n\right)\right)\ \ge\ \frac 1{\max(k\ n)\cdot(\max(k\ n)+1)}$$ There are exactly $\ 2\cdot n+1\ $ points $\ \left(\frac 1r\ \frac 1s\right)\in X\ $ such that $\ \min(r\ s) = n.\ $ For each $\ 2\cdot n+1\ $ of such points $\ x\ $ we have $\ \delta(x)\ge \frac 1{2\cdot n\cdot(2\cdot n+1)}.\ $ Now consider an arbitrary route $\ (p(1)\ p(2)\ \ldots)\ $ of $\ X.\ $ For each of the $\ 2\cdot n+1\ $ points $\ p(t)\ $ such that it is one of points $\ x\ $ from above, we get: $$ d(p(t)\ p(t+1))\ \ge\ \frac 1{2\cdot n\cdot(2\cdot n+1)}$$ Thus the sum of these $\ 2\cdot n + 1\ $ distances is at least $\ \frac 1{2\cdot n}.\ $ Thus (with a harmless abuse of notation): $$\sum_{t=1}^\infty d(p(t)\ p(p(t+1))\ \gt \sum_{n=1}^\infty \frac 1{2\cdot n}\ \ge\ \infty$$ END of PROOF 

