Uncountable Product of Sequentially Compact Spaces is not always Sequentially Compact

Theorem
Let $I$ be an indexing set with uncountable cardinality.

Let $\left\langle{\left({S_\alpha, \tau_\alpha}\right)}\right \rangle_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.

Let $\displaystyle \left({S, \tau}\right) = \prod_{\alpha \mathop \in I} \left({S_\alpha, \tau_\alpha}\right)$ be the product space of $\left\langle{\left({S_\alpha, \tau_\alpha}\right)}\right \rangle_{\alpha \mathop \in I}$.

Let each of $\left({S_\alpha, \tau_\alpha}\right)$ be sequentially compact.

Then it is not necessarily the case that $\left({S, \tau}\right)$ is also sequentially compact.

Proof
Let $\mathbb I$ denote the closed unit interval.

Let $T = \left({\mathbb I, \tau}\right)$ be the topological space consisting of $\mathbb I$ under the usual (Euclidean) topology.

Let $T' = \mathbb I^{\mathbb I} = \left({\displaystyle \prod_{\alpha \mathop \in \mathbb I} \left({\mathbb I, \tau}\right)_\alpha, \tau'}\right)$ be the uncountable Cartesian product of $\left({\mathbb I, \tau}\right)$ indexed by $\mathbb I$ with the Tychonoff topology $\tau'$.

From Closed Real Interval is Sequentially Compact, $T$ is sequentially compact.

But from Uncountable Cartesian Product of Closed Unit Interval is not Sequentially Compact, $T$ is not a sequentially compact space.

Hence the result.