Sub sequence limits Let $\left\{x_{n}\right\}$ be a sequence in the
$((0, 1) \cap \mathbb{Q} ) \subset \mathbb{R}$ with following properties:
1.$\left|x_{\text {n+1}}-x_{n}\right|<\frac{1}{n} ,n \in \mathbb{N}$
2.$\sup \{x_n | n \in \mathbb{N} \}=1$
3.$\inf\{x_n | n \in \mathbb{N} \}=0$

show that  set of subsequential limits of $\left\{x_{n}\right\} $ is $[0,1]$.

It's clear that $\{x_n\}$ is not constant. One of example of this sequence is enumeration of all rational numbers in the $(0,1)$ . Let $E$ is  set of subsequential limits of $\left\{x_{n}\right\} $
Since $\{x_n\}$ is a bounded sequence by the Bolzano–Weierstrass Theorem contain a convergent subsequence, then $E$ is non-empty.
Now for every $c \in [0,1]$ I want to construct sub sequence  $\{r_n\}$ of $\{x_n\}$ such that $\{r_n\}$ is convergence to $c$ .
Let Consider the sequence $\{r_n\}$ such that $r_n \in \{y\in \Bbb R \ | \ |y - c| \lt \frac 1 n\} \cap \{x_n\}$ for each $n \in \Bbb N$ such that $r_n \neq r_i$ for $i \in \{1,2,.. n - 1\}$. Suppose $r_m$ does not exist for some $n \in \Bbb N$. Then either there are no elements of $\{x_n\}$ in $ \{y\in \Bbb R \ | \ |y - c| \lt \frac 1 m\} \implies $ No subsequence in $\{x_n\}$ can converge to a limit in $\{y\in \Bbb R \ | \ |y - c| \lt \frac 1 m\} \implies c - \frac 1 m$ is an upper bound for $E$ leading to a contradiction.
OR
For some $m \in \Bbb N$ Every element in $\{y\in \Bbb R \ | \ |y - c| \lt \frac 1 m\} \cap (x_n)$ is in $\{r_1, r_2,.. r_{m - 1}\}$. Let $M = \text{Max}  \{r_1, r_2,.. r_{m - 1} \}$. If $M \lt c$ then picking $ l = \frac {c - M} 2 \gt 0$ we have that $c - l$ is an upper bound for $E$ leading to a contradiction. If $M \ge c$ then either the elements in $\{r_1, r_2,.. r_{m - 1}\}$ form a subsequence of $(x_n)$ or they do not. If they do then their limit is either equal to $c$ or is greater than it. If they do not form a subsequence of $\{x_n\}$ they are irrelevant.

Is this true? Is there another way to solve this problem ?

in general case how we can prove following lemma ?

lemma : let $(u_n)$ be a bounded sequence such that $u_{n+1}-u_n\to 0$. Then the set of subsequential limits of $(u_n)$ is a closed interval $[\alpha, \beta]$.$\sup \{x_n | n \in \mathbb{N} \}= \beta$
$\inf\{x_n | n \in \mathbb{N} \}=\alpha$

 A: Your argument has several problems. First,
$$r_n\in\left\{y\in\Bbb R:|y-c|<\frac1n\right\}\cap\{x_n\}$$
doesn’t make sense: $\{x_n\}$ is a one-element set, so what you’re written defines $r_n$ to be $x_n$ if $|x_n-c|<\frac1n$ and leaves it undefined otherwise. I suspect that what you mean is to let
$$r_n\in\left(c-\frac1n,c+\frac1n\right)\cap\{x_k:k\in\Bbb N\}\,.\tag{1}$$
Note that you cannot use $n$ as the dummy index in that last set, since $n$ already denotes the specific index of the $r$ term that you’re choosing.
If at some stage $m$ the set in $(1)$ is empty, it’s true that no subsequence of $\langle x_k:k\in\Bbb N\rangle$ can converge to any point in $\left(c-\frac1m,c+\frac1m\right)$, but this does not imply that $c-\frac1m$ is an upper bound for $E$: $E$ may have elements greater than $c+\frac1m$. Thus, at this point you do not have a contradiction.
Your OR alternative, in which
$$\left(c-\frac1n,c+\frac1n\right)\cap\{x_k:k\in\Bbb N\}=\{r_0,\ldots,r_{m-1}\}\,,$$
runs into the same failure to produce a contradiction.
Finally, there is the problem that even if you can use $(1)$ to choose an $r_n$ for each $n\in\Bbb N$, there is no reason to think that $\langle r_n:n\in\Bbb N\rangle$ is a subsequence of $\langle x_n:n\in\Bbb N\rangle$: it might be something like
$$\langle x_{100},x_{666},x_{20},x_{19},x_{18},x_5,x_{420},x_2,\ldots\rangle\,.$$
You would still need to show that it has a subsequence that is a subsequence of $\langle x_n:n\in\Bbb N\rangle$.
Here is an easier approach:
Let $\sigma=\langle x_n:n\in\Bbb N\rangle$, and suppose that $a\in(0,1)$ is not a subsequential limit of $\sigma$. (It’s clear from the second and third conditions that $0$ and $1$ are subsequential limits of $\sigma$.) Then there is an $\epsilon>0$ such that $0<a-\epsilon<a+\epsilon<1$, and for every $n\in\Bbb N$ there is a $k\ge n$ such that $x_k\notin(a-\epsilon,a+\epsilon)$.
There is an $n_0\in\Bbb N$ such that $\frac1{n_0}<2\epsilon$. The second condition ensures that there is a $k\ge n_0$ such that $x_k\ge a+\epsilon$, and the third condition ensures that there is an $n>k$ such that $x_n\le a-\epsilon$. Let $m=\max\{\ell<n:x_\ell\ge a+\epsilon\}$; then $m\ge k\ge n_0$, $x_m\ge a+\epsilon$, and $x_{m+1}\le a-\epsilon$, so $|x_{m+1}-x_m|\ge 2\epsilon>\frac1{n_0}$. But this is impossible: the first condition implies that $|x_{m+1}-x_m|<\frac1m\le\frac1{n_0}$. Thus, no such $a$ exists, and every point of $[0,1]$ is a subsequential limit of $\sigma$.
It is straightforward to modify this argument to prove the more general lemma that you added at the end of the question.
