An example which is homeomorphic to yours, but does not depend on expansion of reals etc: take the index set $I = \{0,1\}^{\mathbb{N}}$, the set of all $0$-$1$-sequences (the Cantor set, to a topologist). This has natural functions $\pi_n$ that sends an $i \in I$ (which is really a sequence, i.e itself a function from $\mathbb{N}$ to $\{0,1\}$) to its $n$-th coordinate; this corresponds to the $n$-th decimal of the real number $i$ in your example.
Now consider $X = \{0,1\}^I$. This is compact by Tychonoff's theorem. Define $f_n \in X$ (for every (fixed for now) $n$) as $f_n(i) = \pi_n(i)$. This is well-defined, because a member of $X$ is just a function from $I$ to $\{0,1\}$, and for some fixed $n$ we have that we send $i \in I$ to its $n$-th coordinate, which is $0$ or $1$. So we have a sequence $(f_n)$ in $X$.
A basic fact for the product topology, for any sequence $(x_n)$ in $X$ (so all $x_n$ are functions from $I$ to $\{0,1\}$!): $(x_n) \rightarrow x (\in X)$ iff for all $i \in I$, $x_n(i) \rightarrow x(i)$. So convergence is determined pointwise. In fact we only need the implication from left to right (which just says that all projections are sequentially continuous).
Suppose that $(f_n)_n$ has a convergent subsequence $f_{n_k}$. Define a point $j \in I$ by $j_{n_{2k}} = 1$ for all $k$, $j_m = 0$ for all other $m$.
Then $f_{n_k}(j) = \pi_{n_k}(j) = j_{n_k}$ which is just (as $k$ goes from $0,1,\ldots$) the sequence $1, 0, 1, 0, 1, 0, \ldots$ by how $j$ was defined. So this subsequence does not converge pointwise at $j$ so cannot converge in $X$ at all. Contradiction.
So $X$ is not sequentially compact.
It turns out that if $X_i$ are non-trivial (more than 1 point) sequentially compact Tychonoff spaces for $i \in I$, then $\prod_i X_i$ is sequentially compact for countable index set $I$, and never sequentially compact for $I$ of size $\mathfrak{c}$. There is some turning point cardinal in between, related to the cardinal invariant called the splitting number. If CH holds, of course all is known, but in general we could have that e.g. $\{0,1\}^{\aleph_2}$ is sequentially compact, while $\{0,1\}^{\aleph_3}$ is not, even with $\aleph_3 < \mathfrak{c}$, etc. This as a set-theory aside and it shows that we need "big" products to kill sequential compactness in products.
According to the Handbook of Set-theoretic Topology (a book that any serious general topologist should have IMO) in the chapters by Vaughn and by van Douwen the fact is mentioned (and partially proved, plus references tot the original papers can be found there as well) that a product of $<\mathfrak{t}$ many sequentially compact (Hausdorff) spaces is again sequentially compact, where $\mathfrak{t}$ is the cardinal invariant known as the tower number (defined on page 115 in the handbook). It is consistent with ZFC that $\mathfrak{c} = \aleph_2 = \mathfrak{t}$ (e.g. by theorem 5.1 of van Douwen's chapter).
In such a model no such $X$ as desired could exist. So the answer is no.
Best Answer
So here is my answer
As you can see this example is too straightforward, and I would like to see other examples.
Thanks.