Question: Let $N,X\_1,\ldots$ be independent random variables with $N\sim \mathrm{Poisson}(\lambda)$ and $X\_k \sim \mathrm{Bernoulli}(1/2)$ for all $k\geq 1.$

Define $Y\_1 = \sum\_{k=1}^N X\_k$ and $Y\_2 = N-Y\_1$. 

If $N=0$ then $Y\_1 = 0$.
How would I show that $Y\_1$ and $Y\_2$ are independent? 

I first found the distribution of $Y\_1$, which is $\mathrm{Poisson}(\frac{\lambda}{2})$. 

However, I'm not sure how to go from here. 

Note that $Y\_1|N \sim\mathrm{Binomial}(N,1/2)$, so
 $$\begin{align}
\mathbb{P}(Y\_1 = y) &= \sum\_{n = y}^\infty \mathbb{P}(Y\_1 = y|N=n) \mathbb{P}(N=n)\\ &= \sum\_{n=y}^\infty \binom{n}{y}(\frac{1}{2})^{n} \times \frac{e^{-\lambda} \lambda^n}{n!} \\
&= \sum\_{n=y}^\infty \frac{1}{(n-y)!y!}(\frac{\lambda}{2})^n \times e^{-\lambda}
\\ &= \frac{e^{-\lambda/2}}{y!}\sum\_{i=0}^\infty \frac{e^{-\lambda/2}(\frac{\lambda}{2})^{y+i}}{i!} \\
&= \frac{(\frac{\lambda}{2})^y e^{-\lambda/2}}{y!} \times 1
\end{align}
$$
since the sum is just the density of a $\mathrm{Poisson}(\frac{\lambda}{2})$ distribution.

Answer: 
Proceeding similarly by conditioning on $N$, you will find that $Y\_2$ is also $\text{Poisson}(\lambda/2)$.

Finally, $P(Y\_1=j,Y\_2=k)=P(Y\_1=j,N-Y\_1=k)$

$\qquad\qquad\qquad\qquad\qquad\quad=P(Y\_1=j,N=k+j)$

$\qquad\qquad\qquad\qquad\qquad\quad=P(Y\_1=j\mid N=k+j)P(N=k+j)$

$\qquad\qquad\qquad\qquad\qquad\quad=\displaystyle\binom{k+j}{j}\left(\frac{\lambda}{2}\right)^{k+j}\frac{e^{-\lambda}}{(k+j)!}$

$\qquad\qquad\qquad\qquad\qquad\quad=\displaystyle\frac{e^{-\lambda}(\frac{\lambda}{2})^{k+j}}{k!j!}\mathbf1\_{{j,k}\{0,1,2,\cdots\}}$

$\qquad\qquad\qquad\qquad\qquad\quad=\displaystyle P(Y\_1=j)P(Y\_2=k)$.
If you are familiar with the Poisson process, there's a more intuitive approach. It's clear that each $Y\_{i}$ is $Poisson(\lambda/2)$: Just imagine their sum as being generated by a Poisson process running for time 1 with force $\lambda$, which clearly gives the same joint $Y\_{i}$ distribution as in the problem. Clearly each $Y\_{i}$ is the counts from a Poisson process running for time 1 with force $\lambda/2$. Now consider $Z\_{i}, i=1,2$ which are the results of Poisson processes running for time $1/2$ (on the intervals $[0,1/2)$ and $[1/2,1]$). They each also are $Poisson(\lambda/2)$ and are clearly independent. But the joint distribution of $N$ and $Z\_{1}$ is the same as the joint distribution of $N$ and $Y\_{1}$, since squishing the interval by a factor of 2 and doubling the force of the process changes nothing. (Yes, you should work out the math behind this last sentence, but the point of this answer is to show why the result is intuitive.) Thus, since $Z\_{1}$ and $N-Z\_{1}$ are independent (independence of non-overlapping time intervals being a basic property of the Poisson process), so are $Y\_{1}$ and $N-Y\_{1}$.