In infinite dimensions, is it possible that convergence of distances to a sequence always implies convergence of that sequence?

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.

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.