Extension to Bolzano–Weierstrass theorem (required property of a convergent subsequence) According to Bolzano–Weierstrass theorem, it says that any bounded sequence in $\mathbb{R}^n$ has a convergent subsequence. 
Assume that $\{\mathbf{x}_\nu\}$ is a given bounded sequence and $\{\nu_k\}$ is the index set of a convergent subsequence. Can we assume that $\lim_{k\rightarrow \infty}\nu_k-\nu_{k-1}<\infty$? (i.e. subsequent iterates are in finite distance of each other.)
If yes, why do you claim that?
If not, is there any requirement on the sequence $\{\mathbf{x}_\nu\}$ by which, existence of such a convergent subsequence can be guaranteed?
Thank you very much in advance. 
 A: Four miscellaneous observations: 

1) Consider $\{x_k\}=\{0, 1, 1, 0,0, 0, 1,1,1, 1, 0,0,0,0, 0,  …\}$.  The only values a convergent subsequence could converge to are 0 or 1, but either way we would need to skip a progressively growing number of indices. 

2) Let $\{X_1, X_2, X_3, …\}$ be a sequence of iid random variables with a continuous CDF function $F_X(x) = P[X\leq x]$ for all $x \in \mathbb{R}$.  
Claim:
With probability 1, $\{X_1, X_2, X_3, …\}$ does not contain a convergent subsequence $X_{n[k]}$ that satisfies $\limsup_{k \rightarrow\infty} (n[k+1]-n[k]) < \infty$. 
Proof:
For any rational numbers $a,b$ such that $a<b$, the law of large numbers tells us that: 
$$ \lim_{n\rightarrow\infty} \frac{1}{n}\sum_{i=1}^n1\{X_i \in [a,b]\} = F_X(b)-F_X(a) \quad (w.p.1) $$
where $1\{X_i \in [a,b]\}$ is an indicator function that is 1 if $X_i \in [a,b]$, and $0$ else. There are only a countably infinite number of rational intervals   $[a,b]$. Hence, with probability 1, the above holds for all rational intervals. 
Now suppose $\{X_n\}_{n=1}^{\infty}$ contains a convergent subsequence of the desired form. So there is a real number $c \in \mathbb{R}$, an integer  $M>0$, and a subsequence $n[k]$ such that $\lim_{k\rightarrow\infty} X_{n[k]} = c$ and $\limsup_{k\rightarrow\infty} (n[k+1]-n[k]) \leq M$.  The latter implies that $n[k+1]-n[k]\leq M$ for all sufficiently large $k$ (since $n[k+1]-n[k]$ is integer-valued).  Fix $\delta>0$ such that 
$$ F_X(c+\delta) - F_X(c-\delta) \leq 1/(2M)$$
This is possible because the $F_X$ function is continuous.  Fix rational numbers $a,b$ such that $a<c<b$ and 
$$ c-\delta \leq a < c < b \leq c + \delta $$
Since $F_X$ is nondecreasing, we get: 
 $$ F_X(b) - F_X(a) \leq F_X(c+\delta)-F_X(c-\delta) \leq 1/(2M) < 1/M \quad (*)$$
The subsequence $\{X_{n[k]}\}$ converges to $c$ and so it is in the interval $[a,b]$ for all sufficiently large $k$. Hence, eventually the 
sequence $\{X_n\}_{n=1}^{\infty}$ will fall into the interval $[a,b]$ at least once every $M$ steps, and so:
$$ \liminf_{n\rightarrow\infty} \frac{1}{n}\sum_{i=1}^n 1\{X_i \in [a,b]\} \geq 1/M > F_X(b)-F_X(a)$$
where the final inequality holds by (*). 
Thus: 
$$ \lim_{n\rightarrow\infty}  \frac{1}{n}\sum_{i=1}^n 1\{X_i \in [a,b]\} \neq F_X(b)-F_X(a) $$
Hence, $[a,b]$ is a rational interval over which the law of large numbers result does not work.  This occurs with probability 0. $\Box$

3) Take the iteration: 
$$ x[k+1]=(1-\gamma[k])x[k] + \gamma[k]r[k] $$
where $x[0]$ is a given real number, $\gamma[k]$ is a given square-summable sequence, and $r[k]$ is random.  
Suppose that $E[(x[k]-r[k])^2] \leq B$ for all $k$, for some constant $B$. Suppose that 
$$E[r[k]|x[0],x[1],...,x[k]]=x[k]$$ 
Then $x[k]$ is a martingale.  Further, we have for all $k$: 
$$ x[k+1]-x[k] = \gamma[k](r[k]-x[k]) $$
Summing over $k\in\{0, ..., n-1\}$ gives: 
$$ x[n] - x[0] = \sum_{k=0}^{n-1} \gamma[k](r[k]-x[k]) $$
Hence: 
$$ E[(x[n]-x[0])^2] = \sum_{k=0}^{n-1}\gamma[k]^2E[(r[k]-x[k])^2] \leq B\sum_{k=0}^{\infty}\gamma[k]^2 $$
where we have used the fact that $E[(r[k]-x[k])(r[i]-x[i])]=0$ for $i\neq k$. 
So $\sup_n E[x[n]^2]$ is bounded.  By the $L^2$-bounded martingale convergence theorem, we know $x[n]$ converges with probability 1. 

4) Consider $\{X_n\}_{n=1}^{\infty}$ iid uniform over $[0,1]$. 
Claim:
With prob 1, we can find a convergent subsequence $\{X_{n[k]}\}$ that converges to $0.5$ and that satisfies $n[k+1]-n[k] \leq k^{1/2}$ for all sufficiently large $k$. 
Proof: 
Define the subsequence as follows: Define $n[1]=1$. For $k\in\{2, 3, 4, ...\}$, greedily select $n[k]$ as the first integer $n>n[k-1]$ for which:
$$X_n \in \left[.5-\frac{1}{2k^{1/4}}, .5+\frac{1}{2k^{1/4}}\right]$$ 
By construction, the $\{X_{n[k]}\}_{k=1}^{\infty}$ subsequence will converge to $0.5$.  We just need to show the distances between $n[k]$ grow sublinearly.  Notice that at stage $k$, the probability of a sample $X_n$ falling in the above interval is $1/k^{1/4}$. 
Define $Wait[k] = n[k+1]-n[k]$. Fix $k \in \{1, 2, 3, \ldots\}$. For simplicity, treat $k^{1/2}$ as an integer. Then:
$$P[Wait[k] > k^{1/2}] = (1-1/k^{1/4})^{k^{1/2}} = \left((1-1/k^{1/4})^{k^{1/4}}\right)^{k^{1/4}} \approx (1/e)^{k^{1/4}} $$
This decays rapidly.  Notice that: 
$$ \sum_{k=1}^{\infty} P[Wait[k]>k^{1/2}] < \infty$$ 
It follows by Borel-Cantelli that, with prob 1, at most a finite number of the events $\{Wait[k]>k^{1/2}\}$ occur. In particular, $Wait[k] \leq k^{1/2}$ for all sufficiently large $k$. $\Box$
