Prove that A is dense in ℝ if for every open set U of ℝ there exists a point of A in U. Let A be a subset of $\mathbb{R}$. Suppose A has the property that for every open set U of $\mathbb{R}$ there exists a point of A in U. Prove that A is dense in $\mathbb{R}$.
Basically we know that A covers the real numbers so any real number must be in the union of the indexed family of sets. 
My solution: Suppose A is not dense in $\mathbb{R}$. Then there exists $x\in\mathbb{R}$ such that for all sequences $x_n$ in A, $x_n$ does not converge to x. Thus, there exists an open subset of $\mathbb{R}$ $I_r(k)$ such that (k-r,k+r) does not contain x. This is a contradiction since A is an open cover of $\mathbb{R}$ and $x\in\mathbb{R}$.
 A: The closest thing to the proof you started is a proof by contrapositive.  That is, we will show that "$A$ is not dense" implies that "for every open $U$..." is false.
Now, we have to understand what it is that we're trying to prove.  In particular, we should note that

not (for every open set $U$ of $ℝ$: there exists a point of $A$ in $U$) =
  There exists an open set $U$ of $\Bbb R$ such that: not (there exists a point of $A$ in $U$)  =
  There exists an open set $U$ of $\Bbb R$ such that: $U$ contains no points of $A$ =
There exists an open set $U$ of $\Bbb R$ such that: $U \cap A = \emptyset$

Now, suppose that $A$ is not dense.  Then, there is some $x \in \Bbb R$ such that no sequence made from the elements of $A$ converges to $x$.  Now, the tricky consequence of this statement is: 

there is some $r > 0$ such that $(x-r,x+r)$ contains no elements of $A$

I would not consider this an "easy" proof.  From there, if we define $U = (x-r,x+r)$, then $U$ is exactly the kind of open set we wanted.  That ends the proof.
Personally, I would have preferred to go through the direct proof here.

Direct Proof: Suppose that $A$ is such that every open set $U$ contains some element of $A$.  Now, pick any $x \in \Bbb R$
Our goal is to prove that there is a sequence $x_n \to x$ for which each $x_n$ is an element of $A$.  In other words, our goal is to build such a sequence.
Here's how we would build the sequence: for each integer $n\geq 1$, $I_{1/n}(x)$ is an open set, and must therefore contain an element of $A$.  For every such integer $n$, pick any element from $I_{1/n}(x) \cap A$, and call that element $x_n$.  The sequence built in this fashion converges to $x$, as you should verify using the definition of convergence.
Since $x$ was an arbitrary element of $\Bbb R$, the same thing works for every $x$, which tells us that $A$ is dense.
