How to use the Lemma of Heine–Borel in order to prove the existence of an irrational number. After learning about the Theorem of Heine–Borel, which states that for intervals on the real line of type $[a,b]$, $\textit{every open cover has a finite subcover}$.  the lecturer gave us a task to think about, which is to prove the existence of an irrational number by using this lemma. he asked us to think about the closed interval $[0,1]$ for example, and how to build for it an open cover (which is based on the rational numbers) s.t. a finite subcover of it, which exists according to lemma, creates a contradiction. (turns out that it doesn't cover all the inteval $[0,1]$.)
do you guys have any hints for how to do that? 
 A: Here is one possibility.  Suppose all the numbers in [0,1] are rational.  We know the rational numbers in [0,1] are countable, i.e. they can be enumerated $\{ q_1, q_2, q_3, \dots \}$.  Enclose $q_1$ in an open interval of width $1/2$, $q_2$ in an open interval of width $1/4$, $q_3$ in an open interval of width $1/8$, etc.  Since we are assuming all the numbers in [0,1] are rational, this is an open cover of [0,1], hence there is a finite subcover.  The subcover has a total width strictly less than $1/2 + 1/4 + 1/8 + \dots = 1$.  But you can't cover [0,1] with a finite set of intervals of total width less than $1$.  Contradiction.
A: This does not quite answer the OP's question because it uses the existence of an irrational to show that $\mathbb{Q}\cap[0,1]$ is not compact.  However, the OP asked to prove existence.
Hint: $\pi-3$ is irrational and within $[0,1]$.  Let $x_n$ be the truncated $\pi-3$ (i.e., the decimal expansion of $x_n$ agrees with $\pi-3$ on the first $n$ positions and is $0$'s after that).  Observe that $x_n$ is rational.  Then the open intervals $(-\infty,x_n)$ cover $[0,\pi-3)$ but there is no finite subcover (requires proof).  Now, think about doing the same thing on the other side of $\pi-3$.
