Modeling the process
Define $\mathcal{S}$ as the set of possible values for the Markov chain:
$$\mathcal{S} = \{0, 0.5, 1, 1.5, …, 9.5, 10\}$$
Note that $S_0=5$ and $S_n \in \mathcal{S}$ for all $n \in \{0, 1, 2, …\}$.
We have
$$S_{n+1} = S_n + A_n \quad \forall n \in \{0, 1, 2, ...\} $$
where
$$ A_n = \left\{ \begin{array}{ll}
(1/2)B_n &\mbox{ if $S_n \notin \{0, 10\}$} \\
0 & \mbox{ otherwise}
\end{array}
\right.$$
where $\{B_n\}_{n=0}^{\infty}$ is an i.i.d. sequence with $P[B_n=1]=P[B_n=-1]=1/2$.
Then
$$\boxed{E[A_n|S_n=s] = 0 \quad, \forall s \in \mathcal{S}} \quad (Eq. 1) $$
Mean
So for each $n \in \{0, 1, 2, ...\}$ we have
\begin{align}
E[S_{n+1}] &\overset{(a)}{=} \sum_{s \in \mathcal{S}}E[S_{n+1}|S_n=s]P[S_n=s] \\
&\overset{(b)}{=} \sum_{s \in \mathcal{S}}E[S_n + A_n|S_n=s]P[S_n=s] \\
&= \sum_{s \in \mathcal{S}}E[s + A_n|S_n=s]P[S_n=s] \\
&= \sum_{s \in \mathcal{S}}(s + E[A_n|S_n=s])P[S_n=s] \\
&\overset{(c)}{=} \sum_{s \in \mathcal{S}}sP[S_n=s] \\
&\overset{(d)}{=} E[S_n]
\end{align}
where (a) holds by the law of total expectation; (b) holds by the fact $S_{n+1}=S_n+A_n$; (c) holds by Eq. (1); (d) holds by definition of expectation.
Since $E[S_0]=5$ we conclude:
$$\boxed{E[S_n]=5 \quad \forall n \in \{0, 1, 2, … \}}$$
Limiting variance
We know $E[S_n]=5$ for all $n$ and so
$$Var(S_n) = E[(S_n-5)^2] = \sum_{s \in \mathcal{S}}(s-5)^2P[S_n=s] $$
Since the process is equally likely to end up at state $0$ or $10$ we have
\begin{align}
\lim_{n\rightarrow\infty} P[S_n=0] &= 1/2\\
\lim_{n\rightarrow\infty} P[S_n=10] &= 1/2\\
\lim_{n\rightarrow\infty} P[S_n=s] &= 0 \quad \forall s \notin \{0, 10\}
\end{align}
so
$$ \boxed{\lim_{n\rightarrow\infty} Var(S_n) = (0-5)^2(1/2) + (10-5)^2(1/2) = 25} $$
Details on variance
Squaring the equation $S_{n+1} = S_n + A_n$ gives
$$S_{n+1}^2 = (S_n+A_n)^2 = S_n^2 + 2S_nA_n + A_n^2 $$
So
$$E[S_{n+1}^2|S_n] = S_n^2 + 2S_nE[A_n|S_n] + E[A_n^2|S_n] = S_n^2 + 0 + (1/4)1_{\{S_n \notin\{0, 10\}\}}$$
where $1_{\{S_n \notin\{0, 10\}\}}$ is an indicator function that is 1 if $S_n \notin \{0,10\}$ and is 0 else. So
$$E[S_{n+1}^2] = E[S_n^2] + (1/4)P[S_n \notin \{0,10\}]$$
Subtracting 25 from both sides gives
$$ Var(S_{n+1}) = Var(S_n) + (1/4)P[S_n \notin \{0,10\}]$$
and $Var(S_0)=0$ so
$$ \boxed{Var(S_n) = (1/4)\sum_{i=0}^{n-1} P[S_i \notin \{0,10\}] \quad \forall n\in \{1, 2, 3, ...\} } $$
Since $P[S_i \notin \{0,10\}] = 1$ for $i \in \{0, 1, 2, 3, ..., 9\}$ we have $$\boxed{Var(S_1)=1/4, Var(S_2)=2/4, Var(S_3) = 3/4, ..., Var(S_{10})= 10/4}$$
On the other hand:
$$ Var(S_{11}) = 10/4 + (1/4)\underbrace{(1-2(1/2)^{10})}_{P[S_{10}\notin\{0,10\}]}$$
In general, the variance increases as $n\rightarrow\infty$ to approach a
limiting value of $25$. It is possible to compute $P[S_i \notin \{0,10\}]$ for all $i$ (for example, by taking powers of a transition probability matrix), but this calculation is more involved.
