This is a community-wiki answer illustrating Iosif Pinelis's solution (with some simplifications).
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Setting. We will write $s_n = \sum_{k=1}^{n} x_k$, so that the recurrence relation takes the form
$$ x_{n+1} = x_n + \frac{n}{s_n}. \tag{RE} $$
Also, let $\alpha$ and $\beta$ by
$$ \alpha = \liminf_{n\to\infty} \frac{x_n}{\sqrt{n}}
\qquad\text{and}\qquad
\beta = \limsup_{n\to\infty} \frac{x_n}{\sqrt{n}}. $$
Our goal is to prove that $\alpha = \beta = \sqrt{3}$. To make use of these quantities, we will frequently utilize the following inequalities:
Lemma. Let $(a_n)$ be any sequence of real numbers, and let $p > 0$. Then
$$ \color{navy}{\liminf_{n\to\infty} \frac{a_n}{n^p}}
\geq \liminf_{n\to\infty} \frac{a_n - a_{n-1}}{pn^{p-1}} \tag{SC1} $$
and
$$ \color{navy}{\limsup_{n\to\infty} \frac{a_n}{n^p}}
\leq \limsup_{n\to\infty} \frac{a_n - a_{n-1}}{pn^{p-1}}. \tag{SC2} $$
Proof. This is an immediate consequence of the Stolz–Cesàro theorem together with the asymptotic formula $n^p - (n-1)^p \sim pn^{p-1}$.
Step 1. Since $(x_n)$ is increasing, we know that $s_n \leq n x_n$. Then by the lemma,
\begin{align*}
\alpha^2
= \liminf_{n\to\infty} \frac{x_n^2}{n}
&\geq \liminf_{n\to\infty} \left( x_{n+1}^2 - x_n^2 \right)
\tag*{by $\color{#2E8B57}{\text{(SC1)}}$} \\
&\geq \liminf_{n\to\infty} \frac{2nx_n}{s_n}
\tag*{by $\color{#2E8B57}{\text{(RE)}}$} \\
&\geq 2.
\tag*{$\because \ s_n \leq n x_n$}
\end{align*}
Now, by the Stolz–Cesàro theorem again,
\begin{align*}
\beta
= \limsup_{n\to\infty} \frac{x_n}{\sqrt{n}}
&\leq \limsup_{n\to\infty} \frac{x_{n+1} - x_n}{\frac{1}{2}n^{-1/2}}
\tag*{by $\color{#2E8B57}{\text{(SC2)}}$} \\
&= \limsup_{n\to\infty} \frac{2n^{3/2}}{s_n}
\tag*{by $\color{#2E8B57}{\text{(RE)}}$} \\
&= \left[ \liminf_{n\to\infty} \frac{s_n}{2n^{3/2}} \right]^{-1} \tag{1} \\
&\leq \left[ \liminf_{n\to\infty} \frac{x_n}{3\sqrt{n}} \right]^{-1}
= \frac{3}{\alpha}.
\tag*{by $\color{#2E8B57}{\text{(SC1)}}$}
\end{align*}
These altogether show that $0 < \alpha \leq \beta < \infty$.
(Remark. Starting from $\alpha$ and applying a similar argument as above, we can show that $\alpha \beta = 3$. However, this does not determine the value of $\alpha$ and $\beta$. So, we will not bother to prove this.)
Step 2. Using the recurrence relation $\color{#2E8B57}{\text{(RE)}}$, we find that
\begin{align*}
s_{n+1}^2 - 2s_n^2 + s_{n-1}^2
&= (s_n + x_{n+1})^2 - 2s_n^2 + (s_n - x_n)^2 \\
&= 2 s_n(x_{n+1} - x_n) + x_{n+1}^2 + x_n^2 \\
&= 2n + x_{n+1}^2 + x_n^2. \tag{2}
\end{align*}
Using this, we get
\begin{align*}
\liminf_{n\to\infty} \frac{s_n^2}{n^3}
&\geq \liminf_{n\to\infty} \frac{s_{n+1}^2 - s_n^2}{3n^2}
\tag*{by $\color{#2E8B57}{\text{(SC1)}}$} \\
&\geq \liminf_{n\to\infty} \frac{s_{n+1}^2 - 2s_n^2 + s_{n+1}^2}{6n}
\tag*{by $\color{#2E8B57}{\text{(SC1)}}$} \\
&\geq \frac{1}{3} + \frac{1}{6} \biggl( \liminf_{n\to\infty} \frac{x_{n+1}^2}{n} \biggr) + \frac{1}{6} \biggl( \liminf_{n\to\infty} \frac{x_n^2}{n} \biggr)
\tag*{by $\color{#2E8B57}{\text{(2)}}$} \\
&= \frac{1 + \alpha^2}{3}.
\end{align*}
Plugging this into $\color{#2E8B57}{\text{(1)}}$,
\begin{align*}
\beta
\leq \left[ \liminf_{n\to\infty} \frac{s_n}{2n^{3/2}} \right]^{-1}
\leq 2\sqrt{\frac{3}{1+\alpha^2}}. \tag{3}
\end{align*}
A similar calculation also shows that
$$ \alpha \geq 2\sqrt{\frac{3}{1+\beta^2}}. \tag{4} $$
Step 3. Define $f(x) = 2\sqrt{\frac{3}{1+x^2}}$, and note that $f$ is decreasing on $[0, \infty)$. So, using $\color{#2E8B57}{\text{(3)}}$ and $\color{#2E8B57}{\text{(4)}}$, we get
$$ \beta \leq f(\alpha) \implies f(f(\alpha)) \leq f(\beta)
\qquad\text{and}\qquad
\alpha \geq f(\beta) \implies f(\alpha) \leq f(f(\beta)) $$
and hence
$$ \beta \leq f(f(\beta)) \qquad\text{and}\qquad f(f(\alpha)) \leq \alpha. $$
It is not hard to check that these inequalities yield $\beta \leq \sqrt{3} \leq \alpha$:
So, using the obvious relation $\alpha \leq \beta$, we conclude that
$$ \sqrt{3} \leq \alpha \leq \beta \leq \sqrt{3} $$
and therefore $\alpha = \beta = \sqrt{3}$.
Best Answer
For any $n \in \mathbb{Z}_{+}$, let $\mathcal{S}_n \subset (0,1]^n$ be the set of non-increasing finite sequences of $n$ terms taking values from $(0,1]$. Let $f_n : \mathcal{S}_n \to \mathbb{R}$ be the function $$\mathcal{S}_n \ni y = (y_1,y_2,\ldots,y_n) \quad\mapsto\quad \sum_{k=1}^n \frac{y_{k-1}^2}{y_k} \quad\text{ where }\quad y_0 = 1 $$ and $M_n = \inf_{y\in \mathcal{S}_n} f_n(y)$.
Notice for any $y = (y_1,\ldots,y_{n+1}) \in \mathcal{S}_{n+1}$, the sequence $z = \left(\frac{y_2}{y_1},\ldots,\frac{y_{n+1}}{y_1}\right) \in \mathcal{S}_n$.
Rewrite $f_{n+1}(y)$ in terms of $y_1$ and $z$, we get
$$f_{n+1}(y) = \frac{1}{y_1} + y_1 f_n(z) \ge \frac{1}{y_1} + y_1 M_n \stackrel{\rm AM \ge GM}{\ge} 2\sqrt{M_n}$$
Taking infimum over $y$, this leads to $$M_{n+1} = \inf_{y\in \mathcal{S}_{n+1}} f_{n+1}(y) \ge 2\sqrt{M_n}$$
It is easy to see $M_1 = 1$. From this, we can deduce
$$M_2 \ge 2\sqrt{M_1} = 2 \implies M_3 \ge 2\sqrt{M_2} = 2^{1+\frac12} \implies \cdots$$ In general, we have
$$M_n \ge L_n \stackrel{def}{=} 2^{1+\frac12+\frac1{2^2} + \cdots + \frac{1}{2^{n-2}}} = 2^{2-2^{2-n}}$$ This lower bound $L_n$ is achievable for any $n$. For $n = 1$, this is obvious.
Let's say for a particular $n$, the lower bound $L_n$ is achieved by sequence $z = (z_1,z_2,\ldots,z_n) \in \mathcal{S}_n$. i.e $f_n(z) = M_n = L_n = 2^{2-2^{2-n}}$.
Consider the sequence $y = \left(\frac{1}{\sqrt{L_n}}, \frac{z_1}{\sqrt{L_n}},\ldots,\frac{z_n}{\sqrt{L_n}}\right)$. It is easy to see $y \in \mathcal{S}_{n+1}$. Furthermore
$$f_{n+1}(y) = \sqrt{L_n} + \frac{1}{\sqrt{L_n}} f_n(z) = 2\sqrt{L_n} = L_{n+1}$$
This means lower bound $L_{n+1}$ is achieved by $y$. By induction, the lower bound $L_n$ is achievable for all $n$. i.e.
$$M_n = \inf_{y\in \mathcal{S}_{n}} f_{n}(y) = \min_{y\in \mathcal{S}_{n}} f_{n}(y) = L_n\quad\text{ for all }\quad n > 0$$
From this, we have $$\lim_{n\to\infty}\inf_{x\in \mathcal{S}_\infty} \left(\frac {x_0^2}{x_1}+ \cdots + \frac{x_{n-1}^2}{ x_n}\right) = \lim_{n\to\infty}\inf_{x\in \mathcal{S}_n} f_n(x) = \lim_{n\to\infty} 2^{2-2^{2-n}} = 4$$