I'll try it for 2D and then we can get 1D as a corollary [excercise!]...
This is the only proof I know of, there may be a more intuitive (and less messy without tex!) proof out there, but I like this one- it uses generating functions in a really nifty way.
Consider the probability of being at the origin after 2n steps (notice we cannot return in an odd number of steps):
$u_0=1$
$u_{2n} = p(n,n,0,0)+ p(n-1,n-1,1,1)+...+p(n-k,n-k,k,k)+...+p(0,0,n,n)$ (when $n\neq0$)
Here $p(u,d,l,r)$ is the probability of the first 2n steps being u up, d down, l left and r right in any order. Each order has probability $\frac{1}{4^{2n}}$, and there are $\frac{(2n)!}{u!d!l!r!}$ distinct orders, giving $p(n-k,n-k,k,k)=\frac{1}{4^{2n}} \frac{(2n)!}{(n-k)!(n-k)!k!k!}$
Now, since $(2n)!=n! n! \binom{2n}{n}$ we have $p(n-k,n-k,k,k)=\frac{1}{4^{2n}} \binom{2n}{n} \binom{n}{k}^2$ giving
$u_{2n}= \frac{1}{4^{2n}} \Sigma_k \binom{2n}{n} \binom{n}{k}^2$ which, by one of those silly binomial results, can be contracted to
$u_{2n}= \frac{1}{4^{2n}} \binom{2n}{n}^2$
Let us put that in our back pocket for now and consider instead the probability of first returning after 2n steps $f_{2n}$ --- this is rather difficult to tackle directly but we can make ourselves a cunning little formula involving it, jazz out some generating function fun and seal the deal with some.
The formula in question is:
$u_{2n}= f_2 u_{2n-2}+ f_4 u_{2n-4} +....+f_{2n-2}u_{2} + f_{2n}$
Which we shall not prove so much as explain: to return to the origin after 2n steps (LHS) you must either first return after 2 steps and do a 'return to the origin in 2n-2 steps walk' (first term RHS) or first return after 4 steps and do a 'return to the origin in 2n-4 steps walk' (2nd term) or... or first return after 2n-2 steps and do a 'return to the origin in 2 steps walk or first return to the origin after 2n steps.
We shall now tweak our formula just a tiny bit so it has the right symmetry properties for what is to come, we do this by adding $f_0=0$ and $u_0=1$ to give
$u_{2n}= f_0 u_{2n}+ f_2 u_{2n-2}+ f_4 u_{2n-4} +....+f_{2n-2}u_{2} + f_{2n}u_0$
Which is secretly a statement about generating functions (this is the sweet bit!), see look at the generating functions of $u_n$, $f_n$:
$U(x)= \Sigma_m u_{2m} x^{2m}$, $F(x)= \Sigma_m f_{2m} x^{2m}$
We see:
$U(x)= 1+ U(x)F(x)$
(where the '1+' is to compensate for the fact that $u_0$ does not appear in the product) Rearranging to:
$F(x)= \frac{U(x)-1}{U(x)}$
Observe that the probability of return is $\Sigma f_n= F(1)$, which comes out as 1 because $U(1)$ diverges by some tedious stirlings formula bounds that I forget.
Edit: Until Tex comes online, this is pretty unreadable, so here's a link to some lecture notes I found with the same proof (and, fortunately, the same notation!). Enjoy!
Edit$^2$: Hooray, Tex has come online!!! Enjoy.