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A friend of mine, in a conversation, said that all lines in $\mathbb{R}^2$ are closed sets under the usual metric. I tried coming up with counter-examples in vain. Is it true?

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    Your friend is correct. – 2017-02-19
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    It is: if you remove a line from the plane, the resulting set is open, that is, every point is in it together with some open ball. – 2017-02-19
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    It's true. And a very good exercise to prove via definitions. – 2017-02-19

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This is true. How you prove it depends exactly on what your definition of "line" is. For instance, if you define a line as a set of the form $$\{(x,y)\in\mathbb{R}^2:ax+by=c\}$$ for some $a,b,c\in\mathbb{R}^2$ with $a$ and $b$ not both $0$, then such a set is closed because it is the inverse image of the closed set $\{c\}\subset\mathbb{R}$ under the continuous map $f:\mathbb{R}^2\to\mathbb{R}$ defined by $f(x,y)=ax+by$.

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    does the line have to be straight? sorry about not making it explicit. the statement, I guess, still works for all lines, straight, curved, closed (as in boundary of closed surfaces. – 2017-02-19
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    The term "line" normally means "straight line". If you allow more general kinds of "lines", the answer depends on how you define "line" (it is unclear to me what precise definition you have in mind). – 2017-02-19
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    I mean like curves. for eg, topologist's sine curve, sine curve, straight lines, a closed path, rectangular hyperbola. these kind of lines. – 2017-02-19
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    You are listing a large number of examples of "curves", but it is not at all obvious how to turn these examples into a precise definition of the term "curve". You can't prove anything about "curves" in general without a precise definition. – 2017-02-19
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    Curves are often defined as continuous images of $[0,1]$ (parametrisations) and thus are compact hence compact. A straight line is not a curve in this sense. Maybe you want continuous images of $\mathbb{R}$, but these not not be closed, even if the function is injective. So deifne your terms first, what is a "curve"? space filling curves historically showed how misguided the definition could be. – 2017-02-19
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Let $L$ be any line in $\mathbb{R}^2$. It suffices to show that $\mathbb{R}^2 \setminus L$ is open. To do this, we need to show that, for any point $p \in \mathbb{R}^2 \setminus L$, there exists a ball of some radius $\delta$ centered at $p$ that is contained entirely inside of $\mathbb{R}^2 \setminus L$.

To do this, we need to show that there is a nonzero distance between $p$ and $L$. Suppose $p = (x_1, y_1)$. Either $L$ is a vertical line $x = c$ for some $c \in \mathbb{R}$, in which case the minimum distance is simply the absolute value of the difference between $c$ and $x_1$. Otherwise, we can parametrize $L$ as $\alpha(t) = (t, mt + b)$ for some $m, b \in \mathbb{R}$. The distance between $p$ and $L$ at some time $t$ is given by $d(t) = \sqrt{(t-x_1)^2 + (mt+b - y_2)^2}$. We want to show that $\inf \{d(t) \ | \ t \in \mathbb{R} \} > 0$. Since $d(t)$ is always non-negative, it suffices to show that $d(t)^2 = (t-x_1)^2 + (mt+b - y_2)^2$ has a nonzero infimum. Finding the infimum is a standard calculus problem.