I would like to show the Bolzano-Weierstrass in $\mathbb{R}^n$, I have seen this theorem in $\mathbb{R}$ and I know it can be shown by induction, something I will try now.
Theorem : Every bounded sequence in $\mathbb{R}^n$ has a convergent subsequence in $\mathbb{R}^n$
Proof : By induction : we know that it is true for $n=1$, we assume this holds for $p=k$ and we want to show this holds for $p=k+1$. Consider $(x_n)$ a bounded sequence in $\mathbb{R}^{k+1}$. For all $n\in\mathbb{N}$ we can write $x_n = (a_n, x_{n}^{k+1})$ where $a_n = (x_{n}^{1}, …, x_{n}^{k})\in\mathbb{R}^{k}$ since $\mathbb{R}^{k+1}$ is isomorphic to $\mathbb{R}^{k}\times\mathbb{R}$ so this rewritting of $x_n$, even if not exactly the same "element" as the initial $x_n$, can be treated as the same.
Clearly, for all $n\in\mathbb{N}, \exists M>0 : \lvert x_{n}^{k+1}\rvert\leq\lVert x_n\rVert_{2}\leq M$ and $\lVert a_n\rVert_{2}\leq\lVert x_n\rVert_{2}\leq M$.
Using the induction hypothesis, we know that $(a_n)$ has a convergent subsquence $(a_{n_j})$, we denote $x\in\mathbb{R}^{k}$ its limit. Now, the sequence $(x_{n}^{k+1})$ has also a convergent subsequence in $\mathbb{R}$ by Bolzano-Weierstrass with limit $x^{k+1}\in\mathbb{R}$. This shows that $x_n$ admits a convergent subsequence $(a_{n_j}, x_{n_j}^{k+1})$ whose limit is $(x,x^{k+1}) = (x^1, …, x^{k+1})\in\mathbb{R}^{k+1}$ so the Bolzano-Weierstrass is true for $p=k+1$, which concludes the proof.
This seems correct to you?
EDIT : My proof is false, we need to take another subsequence to be sure to have a vector with coordinate that have the same index ! Thanks to FShrike, Vercassivelaunos and egreg, you will find three very clear answers to this problem below.
Best Answer
You can fix this without too much bother. Heed the comments well, because they gave you very important warnings. You must make sure you are using the same subsequence for both components, if you want to take a genuine subsequence of $(x_n)_n$ in $\Bbb R^{k+1}$.
By the ordinary BW theorem, there is an increasing sequence $(n_j)_j$ with $(x^{(k+1)}_{n_j})_j$ convergent. Consider the subsequence $(a_{n_j})_j$ which is still a bounded sequence in $\Bbb R^k$. Using the inductive hypothesis, there is a further subsequence $(n_{j_i})_i$ along which $(a_{n_{j_i}})_i$ is convergent. If this is confusing, declare the sequence $b_j:=a_{n_j}$ in $\Bbb R^k$ and note $(b_j)_j$ is a bounded sequence: by induction, it has a convergent subsequence $(b_{j_i})_i$, which translates to $(a_{n_{j_i}})_i$.
Then, because subsequences of a convergent sequence are convergent, we know that $(x^{(k+1)}_{n_{j_i}})_i$ is convergent. It now follows that $(x_{n_{j_i}})_i$ is a convergent subsequence in $\Bbb R^{k+1}$ since this subsequence is convergent in both the $a$ and $x^{(k+1)}$ components.
If you know about this sort of thing, you might like to learn that Bolzano-Weierstrass is true in any complete metric space where bounded subsets are always totally bounded. Sequential compactness (essentially this is Bolzano-Weierstrass) is equivalent to compactness which is further (generalised Heine-Borel) equivalent to completeness and total boundedness (in Euclidean space, that is just closed and bounded).