The sequential compactness of $[0,1]^{\omega_1}$ is undecidable in ZFC: as noted above $[0,1]^{[0,1]}$, so under CH $[0,1]^{\omega_1}$is not sequentially compact; on the other hand $\mathrm{MA}+\neg\mathrm{CH}$ it is sequentially compact. Thus the question is still open.

$\mathrm{MA}$ implies that any product of fewer than continuum many sequentially compact spaces is sequentially compact. In the case of $\aleph_1$ many and when $\mathrm{MA}+\neg\mathrm{CH}$ is assumd you follows the proof for products with countably many factors and produce, given a sequence $(x_n)$ in the product, infinite subsets $A_\alpha$ of $\mathbb{N}$ such that $(x_n)$ restricted to $A_\alpha$ converges on the first $\alpha$ coordinates and such that $A_\alpha\setminus A_\beta$ is finite whenever $\beta<\alpha$. $\mathrm{MA}+\neg\mathrm{CH}$ now implies there is an infinite set $A$ such that $A\setminus A_\alpha$ is finite for all $\alpha$. Then $(x_n)$ restricted to $A$ converges in the full product.

A very nice introduction is still Mary Ellen Rudin's article in the Handbook of Mathematical Logic.

The sequential compactness of $[0,1]^{\omega_1}$ is undecidable in ZFC: as noted above $[0,1]^{[0,1]}$ is not, so under CH $[0,1]^{\omega_1}$is not sequentially compact; on the other hand $\mathrm{MA}+\neg\mathrm{CH}$ it is sequentially compact. Thus the question is still open.

$\mathrm{MA}$ implies that any product of fewer than continuum many sequentially compact spaces is sequentially compact. In the case of $\aleph_1$ many and when $\mathrm{MA}+\neg\mathrm{CH}$ is assumd you follow the proof for products with countably many factors and produce, given a sequence $(x_n)$ in the product, infinite subsets $A_\alpha$ of $\mathbb{N}$ such that $(x_n)$ restricted to $A_\alpha$ converges on the first $\alpha$ coordinates and such that $A_\alpha\setminus A_\beta$ is finite whenever $\beta<\alpha$. $\mathrm{MA}+\neg\mathrm{CH}$ now implies there is an infinite set $A$ such that $A\setminus A_\alpha$ is finite for all $\alpha$. Then $(x_n)$ restricted to $A$ converges in the full product.

A very nice introduction is still Mary Ellen Rudin's article in the Handbook of Mathematical Logic.

The sequential compactness of $[0,1]^{\omega_1}$ is undecidable in ZFC: as noted above $[0,1]^{[0,1]}$, so under CH $[0,1]^{\omega_1}$is not sequentially compact; on the other hand $\mathrm{MA}+\neg\mathrm{CH}$ it is sequentially compact. Thus the question is still open.

The sequential compactness of $[0,1]^{\omega_1}$ is undecidable in ZFC: as noted above $[0,1]^{[0,1]}$, so under CH $[0,1]^{\omega_1}$is not sequentially compact; on the other hand $\mathrm{MA}+\neg\mathrm{CH}$ it is sequentially compact. Thus the question is still open.

$\mathrm{MA}$ implies that any product of fewer than continuum many sequentially compact spaces is sequentially compact. In the case of $\aleph_1$ many and when $\mathrm{MA}+\neg\mathrm{CH}$ is assumd you follows the proof for products with countably many factors and produce, given a sequence $(x_n)$ in the product, infinite subsets $A_\alpha$ of $\mathbb{N}$ such that $(x_n)$ restricted to $A_\alpha$ converges on the first $\alpha$ coordinates and such that $A_\alpha\setminus A_\beta$ is finite whenever $\beta<\alpha$. $\mathrm{MA}+\neg\mathrm{CH}$ now implies there is an infinite set $A$ such that $A\setminus A_\alpha$ is finite for all $\alpha$. Then $(x_n)$ restricted to $A$ converges in the full product.

A very nice introduction is still Mary Ellen Rudin's article in the Handbook of Mathematical Logic.