Any open set in the Hilbert Cube is the union of open subsets of the form $$U_1 \times ... \times U_n \times X_{n+1} \times .... \times X_{n+k} \times...$$ where $X_k := [0, \frac{1}{k}]$ for $k \in \Bbb N$ and where $U_i$ are open in $X_i$, $1 \leq i \leq n$ and $n$ may vary.
I cannot understand the above form of an open set in Hilbert Cube. Please Help!!
This is pretty much by definition. The Hilbert cube is defined to be the Cartesian product of the intervals $[0,1/n]$.
The topology used is the product topology which is defined in that way.
In case that what is confusing in your problem is that they wrote the factors that are equal to $X_i$ are only in the tail, notice that $n$ is any number and that you can still have some $U_i=X_i$ at the beginning.
But suppose you started by defining the topology through a metric. Suppose that you defined the Hilbert cube as the set $\prod_n[0,1/n]$ with the metric
$$d(x,y)=\left(\sum (x_i-y_i)^2\right)^{1/2}$$
Consider a ball $B(x,r)$ with center $x$ and radius $r>0$. Then notice that if $N$ is such that $\sum_{n>N}\frac{1}{n^2}<r$ (which it exists because $\sum \frac{1}{n^2}<\infty$) then $$\prod_{i=0}^{N}\{x_i\}\times\prod_{n>N}[0,1/n]\subset B(x,r)$$
This shows, that starting at some index all the factors of $B(x,r)$ are the whole interval $[0,1/n]$.