In infinite dimensions, is it possible that convergence of distances to a sequence always implies convergence of that sequence? This is a cross-posted on MSE here.
Let $(X,d)$ be a metric space. Say that $x_n\in X$ is a P-sequence if $\lim_{n\rightarrow\infty}d(x_n,y)$ converges for every $y\in X.$ Say that $(X,d)$ is P-complete if every P-sequence converges. Problem 1133 of the College Mathematics Journal (proposed by Kirk Madsen, solved by Eugene Herman) asks you to prove that $$\text{compact}\Longrightarrow\text{P-complete}\Longrightarrow\text{complete}$$ and that none of these implications go both ways. The implications follow by showing that $$\text{sequence}\Longleftarrow\text{P-sequence}\Longleftarrow\text{Cauchy sequence},$$ since a P-sequence (and thus a Cauchy sequence) converges iff it has a convergent subsequence. To give counterexamples to the converses, there are several possible directions. My question specifically involves normed vector spaces (although it is overkill for the original problem).
For any $n\geq 0$, any norm on $\mathbb R^n$ induces a P-complete metric. This distinguishes compactness and P-completeness, since $\mathbb R^n$ obviously isn't compact when $n>0$. To differentiate P-completeness and completeness, we can note that a Hilbert space is P-complete iff it is finite-dimensional (otherwise, we take a non-repeating sequence of vectors from an orthonormal basis and get a P-sequence that doesn't converge). I wonder if other infinite-dimensional normed spaces (necessarily Banach) might be P-complete. But my knowledge of Banach spaces is very limited, so I don't have much intuition about what examples to try. Also, the property of P-completeness (unlike compactness and completeness) is not closed-hereditary, so we can't just try an something by embedding it in a larger example.
Question:  What is an example of an infinite dimensional, P-complete Banach space?
Examples I've tried:

*

*$\ell^p$ spaces for all $1\leq p< \infty$. They are not P-complete, since the sequence $e_n=(0,\dotsc,0,1,0,\dotsc)$ is a P-sequence but not Cauchy. As was pointed out to me in the comments, $\ell^\infty$ is not P-complete, but you need a different sequence as a counterexample.

*$C(X)$ for $X$ compact Hausdorff, first-countable and infinite. There must be an accumulation point $p\in X$. We can take a sequence of bump functions $f_k$ converging (pointwise) to the characteristic function $\chi_p$. For any $g\in C(X)$, we have $\lim d(g,f_k)=\lVert g-\chi_p\rVert_\infty$. Thus $(f_k)$ is a P-sequence that does not converge (uniformly), because the pointwise limit is discontinuous.

 A: It seems to me that you can show that no infinite-dimensional separable Banach space $X$ is P-complete as follows. Pick any bounded separated sequence $\{x_n\}_{n=1}^\infty$ in $X$ and pick a dense sequence $\{y_i\}$ in $X$. Pick a subsequence in $\{x_n\}$ for which $\|x_n-y_1\|$ converges. Then from this subsequence pick further subsequence for which $\|x_n-y_2\|$ converges. So on. After doing this for all $i$, pick a diagonal subsequence $\{x_{n(k)}\}_{k=1}^\infty$ and show that it satisfies the desired conditions.
A: That every Banach space is contained in a $P$-complete Banach space follows immediately from the following
Theorem.
Let $X$ be a Banach space. Then there exists a Banach space $Y$ containing $X$ in which no separated sequence is a $P$-sequence.
Modulo "abstract nonsense", which I will explain later, the theorem follows from the following proposition, which comes from Christian Remling's remark that the unit vector basis $(e_n)$ of $c_0$ is not a $P$-sequence in $\ell_\infty$.
Proposition. Suppose that $(x_n)$ is a normalized basic sequence in a Banach space $X$. Then there is an isometric embedding $S$ from $X$ into $X \oplus_\infty \ell_\infty$ such that no subsequence of $(Sx_n)$ is a $P$-sequence.
Proof: Since $(x_n)$ is normalized and basic and $\ell_\infty$ is $1$-injective, there is $\alpha >0$ and a contraction $T: X \to \ell_\infty$ such that for all $n$, $Tx_n = \alpha e_n$. Define $S$ from $X$ into $X \oplus_\infty \ell_\infty$ by
$Sx := (x,Tx)$. Since $T$ is a contraction, $S$ is an isometric embedding. We show that $(Sx_n)$ does not contain a $P$-convergent subsequence; this is basically Christian's comment. Let $A$ be any infinite set of natural numbers and take an infinite subset $B$ of $A$ so that $A\setminus B$ is also infinite.  Then the distance from $Sx_n$ to $-1_B$ is $1+\alpha$ if $n$ is in $B$ and one otherwise, so $(x_n)_{n\in A}$ is not a $P$-sequence.
Now comes the soft souping up. By iterating the Proposition transfinitely, we get for any Banach space $X$ a superspace $Z$ such that no normalized basic sequence in $X$ is a $P$-sequence in $Z$. Iterate this $\omega_1$ times to get an increasing transfinite sequence $X_\lambda$, $\lambda < \omega_1$, of Banach spaces with $X_1 = X$ so that no normalized basic sequence in $X_\lambda$ is a $P$-sequence in $X_{\lambda+1}$. Let $Y$ be the union of $X_\lambda$ over $\lambda < \omega_1$. Every sequence in $Y$ is in some $X_\lambda$, hence no normalized basic sequence in $Y$ is a $P$-sequence. This property carries over to the completion of $Y$ by the principle of small perturbations.
Now suppose that $Y$ is a Banach space in which no normalized basic sequence is a $P$-sequence. We claim that also no separated sequence in $Y$ is a $P$-sequence.  Certainly no non norm null basic sequence in $Y$ is a $P$-sequence, and $P$-sequences are bounded, so it is enough to consider a general separated sequence $(x_n)$ that is bounded and bounded away from zero. If the sequence has a basic subsequence, we are done. But it is known (and contained, for example, in the book of Albiac and Kalton), that if such an $(x_n)$ has no basic subsequence then it has a subsequence that converges weakly, so without loss of generality we can assume that $x_n - x$ converges weakly to zero but is bounded and bounded away from zero. But then $x_n - x$ has a basic subsequence, hence $x_n - x$ cannot have a $P$-subsequence, whence  neither can $x_n$.
EDIT 7/27/20:
The reduction of the problem to the theorem above is a consequence of things proved, but perhaps not always explicitly stated, in any course that contains an introduction to metric spaces:
Theorem. Let $M$ be a metric spaces. Then one and only one of the following is true.
A. $M$ is totally bounded.
B. $M$ contains a separated sequence.
A corollary is that every sequence in a metric space either contains a Cauchy subsequence or a separated subsequence.
