Irreducibility means that you can get from any state to any state. Here it's easy to check that we can get from state 1 to 2 to 3 to 4 to 5 to 6 to 1: the transition probabilities, in that order, are $0.5,0.25,0.5,0.25$, and $0.25$, appearing on the first superdiagonal and in the lower lefthand corner. Here's a graph of the transitions; each goes both ways, so I didn't have to worry about arrows.
1 3 |\ /| | \ / | | \ / | | 2 | | / \ | | / \ | |/ \| 6-------4 \ / \ / \ / 5
From it you can easily check that no matter where you start, you can return to that state in either $2$ or $3$ steps; since $\gcd(2,3)=1$, the chain is aperiodic.
To show that the chain is reversible, you must find a probability distribution $\pi=\langle\pi_1,\pi_2,\pi_3,\pi_4,\pi_5,\pi_6\rangle$ such that $\pi_ip_{ij}=\pi_jp_{ji}$ for $1\le i,j\le 6$. Clearly the equations with $i=j$ are satisfied no matter what $\pi_i$ is, so we can ignore them. We can also ignore any pair for which $p_{ij}=p_{ji}=0$, since $0=0$ gives no information about $\pi$. That leaves the following system of equations:
$\begin{array}{} 0.5\pi_1=0.25\pi_2&&0.5\pi_1=0.25\pi_6\\ 0.25\pi_2=0.5\pi_3&&0.25\pi_2=0.25\pi_4&&0.25\pi_2=0.25\pi_6\\ 0.5\pi_3=0.25\pi_4\\ 0.25\pi_4=0.5\pi_5&&0.25\pi_4=0.25\pi_6\\ 0.5\pi_5=0.25\pi_6 \end{array}$
After the fractions are cleared, we have this system:
$\begin{array}{} 2\pi_1=\pi_2&&2\pi_1=\pi_6\\ \pi_2=2\pi_3&&\pi_2=\pi_4&&\pi_2=\pi_6\\ 2\pi_3=\pi_4\\ \pi_4=2\pi_5&&\pi_4=\pi_6\\ 2\pi_5=\pi_6 \end{array}$
Clearly $\pi_2=\pi_4=\pi_6=2\pi_1=2\pi_3=2\pi_5$. Finally, we know that $\sum_{i=1}^6\pi_i=1$, so $1=\sum_{i=1}^6\pi_i=3\pi_1+3\pi_2=9\pi_2\;,$ and therefore $\pi_1=\pi_3=\pi_5=\frac19\text{ and }\pi_2=\pi_4=\pi_6=\frac29\;.$ In other words, the distribution $\pi=\frac19\langle 1,2,1,2,1,2\rangle$ satisfies the detailed balance condition and is therefore reversible. Of course this distribution is stationary.