Functional Analysis: countable Product of Compact Subsets of Metric Spaces is compact

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I'm revising for Functional Analysis exam and I'm trying to solve this problem. The part a) is clear for me but I have problem with b) and c) and hope someone can give me a hint how to start/solve it. I think that in part c) the theorem of Tychonoff plays a role. But here we have compact SUBSET of $X_n$, so I'm not sure if it's even allowed or how I must argue that I can use the theorem. For b) I don't have an idea at all. If someone could give me hint how to start that would be great.

The problem: Let $(X_n,d_n)$,$n \in \mathbb N$, be a family of metric spaces and X the product space $X=\prod_n X_n$. Define $d: X \times X \rightarrow \mathbb R^+$ for $x=(x_n)_{n\in N}, y=(y_n)_{n\in N} \in X$ as follows: $\sum_{n \in I} \frac{1}{2^n} \frac{d_n(x_n,y_n)}{1+d_n(x_n,y_n)}$.

Prove:

a) d is a metric on X. --> I have already done this.

b) Any open neighborhood $U \subset X$ of an arbitrary point $x \in X$ contains a neighborhood of thr form $\prod_{1\le n\le N} U_n(x_n) \times \prod_{n \gt N} X_n$, where $N \in \mathbb N$ and for each $1\le n\le N,U_n(x_n)$ denotes an open neighborhood of $x_n in X_n$.

c) If for any $n \in \mathbb N, K_n \subset X_n$ is compact, then $\prod_{n \in N} K_n$ is a compact subset of X.

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As you are dealing with metric spaces you may instead concentrate on sequential compactness. For this it suffices to take any sequence $(\xi_k)_{k\geq 1}$ in the product $K=\prod_n K_n$ and use the socalled 'diagonal' trick to extract a subsequence that converges in $K$.