Consider a sequence $0, 1, 2, 1, 2,···$. Explain why any sequence ${x_k}$ with the property $x_k =$ $1$ or $ 2$ for each $k$ , is a subsequence of this sequence.
I have tried to use the Bolzano-Weierstrass theorem but I am not sure if it is suitable for this question or not.
On
The question asks to explain this so I'm going to stick with an informal explanation (see the other answer for a more formal proof). Given any sequence of $1$'s and $2$'s let's see how we would go about to extract this as a subsequence of $a_k = \{0,1,2,1,2,1,2,\ldots\}$ (a subsequence is defined as a sequence on the form $\{a_{i_k}\}_{k=1}^\infty$ where $i_k$ is an increasing sequence of integers).
Lets explain the method with a sequence that starts off like $x_k = \color{red}{2},\color{blue}{2},\color{green}{1},\ldots$. We start with the first element $x_1 = \color{red}{2}$. We now scan $a_k$ from the beginning until we find a $2$. The index this corresponds to is $i_1 = 3$: $$a_k = \{0,1,\color{red}{2},1,2,1,2,\ldots\}$$
Next number is $x_2 = \color{blue}{2}$. We start from $k=i_1+1 = 4$ and scan forward till we find another $2$. The index this corresponds to is seen to be $i_2 = 5$: $$a_k = \{0,1,\color{red}{2},1,\color{blue}{2},1,2,\ldots\}$$ Next number is $x_3 = \color{green}{1}$. We starts from $k = i_2+1=6$ and scan forward till we find a $1$. The index this corresponds to is seen to be $i_3 = 6$: $$a_k = \{0,1,\color{red}{2},1,\color{blue}{2},\color{green}{1},2,\ldots\}$$ We continue this procedure to generate the sequence $i_k = \{3,5,6,\ldots\}$ which will have the property, by construction, that $a_{i_k} = x_k$.
This is a simple method to generate the desired subsequence. Now we need to show that it will always work. What might go wrong is that at some point there are not more $1$'s or $2$'s left in the sequence so our search will never find a match. This will never happen here. If the last number we picked was a $1$ then the next two numbers in $a_k$ is $2,1$ and likewise if the last number we picked was a $2$ then the next two numbers in $a_k$ is $1,2$ so no matter what number we have next in $x_k$ it will always be found to the right of the previous element we picked in $a_k$.
To see if you understand it try to show that the specific form of $a_k$ is of no importance here and that all we need for this method to always work is that it has infinite numbers of $1$'s and $2$'s. Also try to convince yourself that you can construct an infinite number of different subsequences (different choices for the $i_k$'s) that matches $x_k$.
Bolzano-Weierstrass cannot help you because the given sequence is not convergent and we are not asked about always non-convergent subsequences.
We start with the sequence $(a_i)_{i=0}^\infty$: \begin{align*} a_0 &= 0 \\ a_i &= \begin{cases} 1 ,& \text{$i$ is odd} \\ 2 ,& \text{$i>0$ is even} \end{cases} \end{align*} and wish to show $(x_k)_{k=1}^\infty$ is a subsequence of $(a_i)$ where $x_k \in \{1,2\}$ for all $k$.
We must find a function $i(k)$ such that $a_{i(k)} = x_k$ for all $k \in [1,\infty)$. Set $i(1) = x_1$. For all $k > 1$, set $$ \Delta i(k) = \begin{cases} 1 ,& x_{k+1} \neq x_k \\ 2 ,& x_{k+1} = x_k \end{cases} \text{.} $$ and then $i(k) = i(k-1) + \Delta i(k)$.
Now we proceed by induction. First, $$ x_1 = a_{i(1)} = a_{x(1)} = \begin{cases} 1 ,& x_1 = 1 \\ 2 ,& x_1 = 2 \end{cases} \text{.} $$ Now suppose $a_{i(k-1} = x_{k-1}$. Then \begin{align*} a_{i(k)} &= a_{i(k-1) + \Delta i(k)} \\ &= a_{i(k-1) + \begin{cases} 1 ,& x_{k} \neq x_{k-1} \\ 2 ,& x_{k} = x_{k-1} \end{cases} } \\ &= a_{\begin{cases} i(k-1) + 1 ,& x_{k} \neq x_{k-1} \\ i(k-1) + 2 ,& x_{k} = x_{k-1} \end{cases} } \\ &= \begin{cases} a_{i(k-1) + 1} ,& x_{k} \neq x_{k-1} \\ a_{i(k-1) + 2} ,& x_{k} = x_{k-1} \end{cases} \\ &= \begin{cases} a_{i(k-1) + 1} ,& x_{k} \neq x_{k-1} \\ a_{i(k-1)} ,& x_{k} = x_{k-1} \end{cases} \text{,} \end{align*} where the last equality uses that adding two to an even number yields an even number and likewise for an odd number. But we are done. If $x_{k} = x_{k-1}$, we have proven $a_{i(k)} = a_{i(k-1)}$, as needed, and if $x_k \neq x_{k-1}$ we switch $a_{i(k)}$ between the even index and odd index subsequences of $a_i$ so we switch $a_{i(k)}$ between $1$ and $2$ precisely when $x_k$ switches. Therefore, we have shown that $(x_k)$ is a subsequence of $(a_i)$.