Tychonoff's theorem for $[0,1]^\mathbb{R}$ According to Tychonoff's theorem any uncountable product of compact spaces is compact with respect to product topology.
Then $[0,1]^\mathbb{R}$, the space of all functions defined on $\mathbb{R}$ taking values in $[0,1]$ is compact w.r.t. the product topology.
Consider the function $\delta_0(x)=\max(0,\min(x,1))$ on real numbers and $\delta_t(x)=\delta_0(x-t)$. For $t\to\infty$, it seems that there is no convergent sub-sequence and again it seems that  $[0,1]^\mathbb{R}$ is then not compact or sequentially compact?
Could someone point out the problem I have? Thanks.

The product topology is the topology of pointwise convergence; your functions converge pointwise to the zero function. 
You are correct that $[0, 1]^{\mathbb{R}}$ is not sequentially compact; I believe $f_n(x) = |\sin nx|$ is an explicit counterexample, but I haven't checked it carefully. But neither sequential compactness nor compactness imply the other in general (and in particular $[0, 1]^{\mathbb{R}}$ is not metrizable). 
