Is a finite subset of a complete metric space again a complete metric space?

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The space $(\mathbb{R}^2, d)$ where $d(x,y)=max \{|x_1-x_2|, |y_1, y_2|\}$ for $x=(x_1, y_1)$ and $y=(x_2, y_2)$ $\in X$ is a complete metric space.

Let $X=\{(0,0), (-\frac{1}{8}, 0), (0, -\frac{1}{5})\}$ with the same distance function defined above.

I have shown that $(X,d)$ is a metric space by brute force since I only have 3 elements in the set (listing all the conditions and showing it).

My problem is how will I show that this subset is a complete metric space. How am I going to construct the Cauchy sequences. Or do you have other solution than this?

Thanks a lot.

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The easiest solution is probably to note that $X$ is finite, thus closed in $\mathbb{R}^2$. We then use that a subset of a complete metric space is complete if and only if it is closed.