Prove that the family of open sets in $\mathbb{R}$ has cardinality equal to $2^{\aleph_0}$ Let $\mathcal{T}$ be the family of all open sets in $\mathbb{R}$. Show that $| \mathcal{T}|=2^{\aleph_0}$
$\textbf{My Attempt:}$ 
I know that $\forall A \in \mathcal{T}$. $A$ is the countable union of open intervals with rational end points. 
I want to use the Cantor-Bernstein Theorem. That is I need to find injective functions $f$ and $g$ such that $f: 2^{\aleph_0} \to \mathcal{T}$ and $g: \mathcal{T} \to 2^{\aleph_0}$.
I know each $A \in \mathcal{T}$ is of the form $A = \bigcup_{x \in A} (r_x,s_x)$ where $r_x,s_x \in \mathbb{Q}$. How can I use this fact to find the injective function $f$ and $g$?
 A: It is often useful to replace $2^{\aleph_0}$ with another set of the same cardinality for the purpose of finding injective functions; I'll do so frequently, and I recommend you learn to think about sets in this way. 
For one direction, finding an injective function $\mathbb R\to\mathcal T$ should be easy. I leave it to you. (Why does $|\mathbb R|=|2^{\aleph_0}|$?)
The other direction is harder. Your observation is not quite right: I don't see how you're associating $x$ to $r_x$, since $x$ could be irrational. I think the fact you're referring to -- a version of it, at least -- is the following.

Let $A\subseteq\mathbb R$ be open. For each rational number $q\in A$ there exists a rational radius $r_q\in\mathbb Q$ such that $(q-r_q,q+r_q)\subseteq A$. Then $A=\bigcup_{q\in A}(q-r_q,q+r_q)$.

Since $A$ is completely characterized by the set $\{(q,r_q)\,:\,q\in A\cap\mathbb Q\}$
(here I mean $(q,r_q)$ as an ordered pair, not an interval!), why don't we make a function $\mathcal T\to 2^{\mathbb Q\times\mathbb Q}$ given by
$A\mapsto \{(q,r_q)\,:\,q\in A\cap\mathbb Q\}$. The map is injective (why?) and $|2^{\mathbb Q\times\mathbb Q}|=|2^{\aleph_0}|$ (why?). So we're done!
Edit: The statement I gave is not correct as written, as user48481 Mirko Swirko pointed out. The problem is that if I pick the $r_q$ to be too small then I might not get all the points in the open set. For instance, if $A=(0,2)$ and I choose $r_q$ so that $\sqrt2\notin(q-r_q,q+r_q)$ for any $q$, which is possible because $\sqrt2$ is irrational, then the union of the $(q-r_q,q+r_q)$ will miss the point $\sqrt2$.
A corrected version of the statement goes like this.

Let $A\subseteq\mathbb R$ be open. There exists an assignment $q\mapsto r_q$ of rational numbers to rational radii (so $q,r_q\in\mathbb Q$) such that $A=\bigcup_{q\in A} (q-r_q,q+r_q)$.

To prove the statement, it suffices to consider the case where $A$ is an interval $(a,b)$, since every open subset of $\mathbb R$ is a countable union of such intervals, as you observed. You can do this by letting $(q_n)_{n=1}^\infty$ being an enumeration of the rationals in $(a,b)$ and choosing $r_{q_n}$ so that, e.g., if $q_n$ is closer to $a$ than $b$ then $q_n-r_{q_n}-a<2^{-n}$.
A: On the one hand, it is clear that $|\mathbb{R}|\le|\mathcal{T}|$, since for each $r\in\mathbb{R}$, we can consider the open set $(r,+\infty)$; it is clear that this assignation is injective.
On the other hand, for each open subset $A\subseteq\mathbb{R}$ there exists an assignation $q\longmapsto r_q$ between rational numbers with $r_q>0$ such that $A=\bigcup_{q\in A\cap\mathbb{Q}}(q-r_q,q+r_q)$ (because the set of open intervals with rational numbers as their extremes is a basis of the usual topology on $\mathbb{R}$).
Since $A$ is completely characterised by the set $\{\langle q,r_q\rangle|\;q\in A\cap\mathbb{Q}\}\subseteq\mathbb{Q}\times(\mathbb{Q}\setminus\{0\})$, the function $f:\mathcal{T}\longrightarrow^{\mathbb{Q}}\mathbb{Q}$ defined by: for each $A\in\mathcal{T}$:
$$f(A)=\{\langle q,r_q\rangle|\;q\in A\cap\mathbb{Q}\}\cup\{\langle q,0\rangle|\;q\not\in A\cap\mathbb{Q}\}$$
Is clearly injective, according to our previous observation, and thus $|\mathcal{T}|\le|^{\mathbb{Q}}\mathbb{Q}|=|\mathbb{R}|$
From the Cantor-Bernstein theorem we conclude that $|\mathcal{T}|=|\mathbb{R}|=2^{\aleph_0}$
A: Late to the party, but I'd like to offer a somewhat different, quicker approach. The idea is taken from a proof regarding analytic sets, in Donald L. Cohn's very readable book on measure theory. Let $\{U_n\}_{n\in \mathbb N}$ be an enumeration of any countable basis for the topology of $\mathbb R$ and let $\mathscr N=\mathbb N^{\mathbb N},$ the collection of sequences of integers. It is not hard to show that $|\mathbb N^{\mathbb N}|=|\mathbb R|.$
Consider the set  $U=\{((n_1,n_2,\cdots,),x):\exists k\  x\in U_{n_k}\}.$ Fix $\textbf n:=(n_1,n_2,\cdots).$ Then the section $U_{\textbf n}=\{x:(\textbf n,x)\in U\}$ is equal to $\{x:\exists k\ x\in U_{n_k}\},$ and so $U_{\textbf n}=\bigcup _kU_{n_k}.$ But an arbitrary set in $\mathbb R$ is open if and only if it has the form $\bigcup _kU_{n_k}$ for some sequence $\textbf n.$
We conclude that a set is open if and only if it is a section of $U,$ which is to say: $\tau_{\mathbb R}=\{U_{\textbf n}\}_{\textbf n\in \mathscr N}.$ It follows that  $|\tau_{\mathbb R}|=|\{U_{\textbf n}\}_{\textbf n\in \mathscr N}|=|\mathbb N^{\mathbb N}|=|\mathbb R|.$
