Several answers point out the following:
The space of irrationals is homeomorphic to the Baire space $\mathbb N^{\mathbb N}$ of functions from $\mathbb N$ to $\mathbb N$.
No one gave an explicit homeomorphism. [PS: It's now pointed out that the answer by Richard Borcherds did so, if very tersely. I skimmed to fast.]
Let $a_1,a_2,a_3,\ldots$ be a sequence of positive integers. Then $$ a_1 + \cfrac 1 {a_2 + \cfrac 1 {a_3 + \cfrac 1 {a_4 +\ddots}}} \in \mathbb R \smallsetminus\mathbb Q. $$ This number cannot be rational since an expansion of a rational number in this way must terminate because of well-ordering of $\mathbb N.$
That gives you the positive irrationals. To see that that is homeomorphic to the space of all irrationals, recall a fact proved by Georg Cantor: any two countable densely linearly ordered sets without endpoints are order-isomorphic to each other. ("Densely" means only that between any two elements there is another (so no knowledge of topology is needed to understand that word).) And an order isomorphism of those two sets of rationals will give you an order isomorphism of those two sets of irrationals.