Proving the set of irrational numbers in $\mathbb{R}$ has the same cardinality as $\mathbb{R}$ Not going to show my whole proof here, but this logic work?

We are showing |R-Q| = |R| => |I| = |R| 
Since I is a subset of R, |I| must be less than or equal to |R|. 
So all we have to show is |I| is not less than |R|.

So started by proving R is uncountable using diagonalization and a contradiction.
Then proved I is uncountable using a contradiction and the fact that I union Q = R. 

Then concluded that since I is uncountable and R is uncountable, then|I| cannot be &lt; |R| => |I| = |R|. 

Does that make sense?
Just show that there are injective functions from one set to another and vice versa. For one, just take the inclusion. For the other send R\Q in (0,1)\Q and Q to (1,2)\Q for example. ( I'm using the fact the the open interval has a bijection with R and that injections preserve cardinality). Then the result follows by the Cantor-Bernstein-Schroeder theorem.
That $\Bbb R\setminus\Bbb Q$and $\Bbb R$ are both uncountable doesn't imply that they have the same cardinality.  There are infinitely many uncountable cardinals, by Cantor's diagonal argument.

Idea: $\Bbb R=(\Bbb R\setminus\Bbb Q)\cup\Bbb Q$.

But, as you noted, $\Bbb R$ is uncountable.

Presumably you know $\Bbb Q$ is countable.

Hence $\Bbb R\setminus\Bbb Q$ must be of the same cardinality as $\Bbb R$.