Question: I need to prove (using the mean value theorem) that for all $x\in(0,\infty)$, the following inequality holds:

$\exp(-x)\geq 1 - x$

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I don't know how the mean value theorem is applicable here, since we don't have a closed interval.

How do I prove the statement?

Answer: 
Hint: let $f(t)=e^{-t}$, and (for a fixed $x$) try using the mean value theorem on the interval $[0,x]$.
Suppose $x>0$. Then the mean value theorem applied to the interval $[0,x]$ says that
$$
\frac{e^{-x}-e^{-0}}{x-0}=-e^{-c}
$$
for some $c\in(0,x)$. Since $c>0$, we have $-c<0$, so $e^{-c}<1$ and so $-e^{-c}>-1$.

Therefore
$$
e^{-x}-1 > -x
$$