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Let $A\subset\mathbb R^2$ be bounded. Then, by definition, there exists $x\in \mathbb R^2$ and $\epsilon >0$ such that for all $a\in A$, $d(x,a)<\epsilon$.

I want to show that for every $\epsilon >0$, there exist a finite number of points $x_0, x_1,\ldots, x_n \in A$ such that $\inf_i d(x_i,x)<\epsilon$ for all $x \in A$.

How do I get there?

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    You want to show that if $A$ is contained in some open disc, then for every $\epsilon$, the set $A$ can be covered by discs of radius $\epsilon$. Well, if $A$ is contained in an open disc, then $A$ is contained in an open square. Let $L$ denote the side length of that square. How many discs of radius $\epsilon$ are needed to cover an $L$-by-$L$ square? (Draw a picture.) – 2017-01-18
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    @avs Only a finite number of these discs are needed, correct? – 2017-01-18
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    That's correct! And, as the square contains $A$, those finitely many discs will end up covering $A$. (As a further step, you might think about generalizing this solution to the case of $\mathbb{R}^{n}$.:) – 2017-01-18

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if $A$ is bounded then its closure $C$ is also bounded.

Notice that $C$ is compact, so proving it is totally bounded is easy, notice that the balls with radius $\epsilon$ centered at points in $C$ cover $C$, by compactness you need only take a finite amount of these balls.

Since $A\subset C$ and $C$ is totally bounded we conclude $A$ is totally bounded.

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    for metric spaces we have compact if and only if closed and totally bounded, and for $\mathbb R^n$ we have compact if and only if closed and totally bounded. So your theorem fits inside the Heine Borel theorem. – 2017-01-18
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    Fernandez Hidalgo, Unfortunately, your suggestion uses the fact that, in a Euclidean space, compact is equivalent to (closed and bounded). What if her exercise is a step toward *obtaining* this fact? – 2017-01-18
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    @avs good point, it might be useful to add an answer for that case. – 2017-01-18
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    @JorgeFernándezHidalgo We have not yet established that compactness is equivalent to closed and totally bounded. – 2017-01-18
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Hint: Notice that $\bar{A}$ must be compact (Heine-Borel theorem) and then consider the open covering $\{B(x,\varepsilon):x\in A\}$ of $\bar{A}$.