I'm trying to show that the Hilbert cube is compact and want to know if I'm taking the right approach. My text defines the Hilbert cube as: $H=\{(x_1,x_2,...) \in [0,1]^{\infty} : for \ each \ n \in \mathbb{N}, |x_n|\leq \dfrac{1}{2^n}\}$
I need to show that it is compact with respect to the metric:
$d(x,y)=\underset{n}{\textrm {sup}}|x_n - y_n|$
In order for the sequence in $H$ to converge, we need each of it's components to converge.
Since, each component $x_n$ is bounded (by $0$ and $\dfrac{1}{2^n}$) for any point in $H$, a sequence (in $H$) of such points would form a sequence $(x_n)$ for each component $n$ and each of those sequences would be bounded.
Therefore, by Bolzano-Weierstrass, they have convergent subsequences.
We can show that for each sequence of points in $H$, we have a subsequence of the sequence of $x_1$s that converges, say ${x_1}_k$, we then have a sequence $({x_1}_k, {x_2}_k,...)$.
For the sequence of ${x_2}_k$s, there exists a convergent subsequence ${{{x_2}_k}_j}$. Since all subsequences of a convergent subsequence also converge, we have that both $({{{x_1}_k}_j})$ and $({{{x_2}_k}_j})$ converges in $({{{x_1}_k}_j}, {{{x_2}_k}_j},...)$.
Repeating this process ad infinitum, we get convergent subsequences for all $x_n$ so that the sequence in $H$ has a subsequence that converges as well and, therefore, $H$ is compact.
Is this correct? If it is, is that enough to show that $H$ is compact under the given supremum metric?
No. This is not correct. As an aside, your paragraph
is so poorly constructed, I could not figure out what even the intent was until after I had read through the rest of the the post and had enough evidence to reconstruct it.
But the actual error in the argument is this: There is not necessarily a relation between the convergent subsequences in the different indices.
For example, the convergent subsequence chosen for $(x_{1n})$ may be $(x_{1(2k)})$, while the convergent subsequence chosen for $(x_{2n})$ may be $(x_{2(2k+1)})$. Thus the convergent subsequences for your first two indices would not ever come from the same points. There is no subsequence $(x_{n_k})$ of $(x_n)$ itself such that $\pi_1(x_{n_k})$ is the $k^{th}$ in your $(x_{1n})$ subsequence and $\pi_2(x_{n_k})$ is the $k^{th}$ in your $(x_{2n})$ subsequence.
You could try to work around this as follows:
Let $n^1_k$ define a subsequence of $(x_n)$ such that $\pi_1(x_{n^1_k})$ converges. Let $n^2_k$ define a subsequence of $(x_{n^1_k})$ such that $\pi_2(x_{n^2_k})$ converges. Etc.
But this still doesn't work. While $\pi_j(x_{n^m_k})$ is guaranteed to converge for each $j \le m$, there is no guarantee of a sequence that works for all indices. For example, it could be that $n^m_1 = m$ for all $m$. Thus any subsequence you tried for the points themselves would start out less than an infinite number of your index-wise convergent subsequences.
Since convergence of sequences depends on the tails of the sequences, you might by careful argument find a way past those obstacles, but it would be tricky to do so. I suggest you try to prove this by other means than sequential compactness.
Instead, I suggest trying to prove that $H$ is complete and totally bounded.