According to my lecturer:
$\text{Cov}(u_t,u_{t-1})=\rho(\sigma_u)^2$
where
$u_t=\rho u_{t-1}+\epsilon_t$
We have been given no more information than this. I would like to prove and understand this result, but to be honest, I do not even know where to start. Hoping someone can point me in the right direction.
Without some extra assumptions the stated result is not necessarily true. In such models it is typically assumed that $\ \epsilon_t\ $ are independent and identically distributed (commonly with a normal distribution) with zero mean, constant variance, and are independent of $\ u_0\ $. These assumptions will imply that $\ \epsilon_t\ $ and $\ u_{t-1}\ $ are uncorrelated, which is all you really need to show that $\ \text{Cov}\big(u_t,u_{t-1}\big)=\rho\sigma_{u_{t-1}}^2\ $. However, the notation $\ \sigma_u\ $ on the right side of the stated result suggests that your lecturer is taking the variance of $\ u_t\ $ to be constant. This cannot be the case unless $\ |\rho|<1\ $ and $\ \epsilon_t\ $ has constant variance. If these conditions do hold, then $\ \sigma_{u_t}^2\rightarrow\frac{\sigma_\epsilon^2}{1-\rho^2}\ $ as $\ t\rightarrow\infty\ $, and if $\ \sigma_{u_0}^2=\frac{\sigma_\epsilon^2}{1-\rho^2}\ $ then $\ u_t\ $ will have that same variance for all $\ t\ $.
Let \begin{align} e_t&=\mathbb{E}\big(u_t\big)\ \ \text{ and}\\ m_t&=\mathbb{E}\big(\epsilon_t\big)\ . \end{align} Then taking expectations of both sides of the equation $$ u_t=\rho u_{t-1}+\epsilon_t $$ gives $$ e_t=\rho e_{t-1}+m_t\ , $$ and subtracting this latter equation from the former gives $$ (1)\hspace{5em}u_t-e_t=\rho\big(u_{t-1}-e_{t-1}\big)+\epsilon_t-m_t\ . $$ If we now multiply this equation by $\ u_{t-1}-e_{t-1}\ $ and take expectations of the result, we get $$ \text{Cov}\big(u_t,u_{t-1}\big)=\rho\sigma_{u_{t-1}}^2+\text{Cov}\big(u_{t-1},\epsilon_t\big)\ , $$ which reduces to the equation given by your lecturer when $\ u_t\ $ has constant variance and $\ u_{t-1},\epsilon_t\ $ are uncorrelated.
For the results on the variance of $\ u_t\ $, square both sides of equation $(1)$ and take expectations to obtain $$ \sigma_{u_t}^2=\rho^2\sigma_{u_{t-1}}^2+2\rho\,\text{Cov}\big(u_{t-1},\epsilon_t\big)+\sigma_{\epsilon_t}^2\ , $$ which reduces to $$ \sigma_{u_t}^2=\rho^2\sigma_{u_{t-1}}^2+\sigma_\epsilon^2 $$ when $\ u_{t-1},\epsilon_t\ $ are uncorrelated and $\ \epsilon_t\ $ has constant variance.
If $\ \rho=\pm1\ $, it follows by induction from this recurrence that $\ \sigma_{u_t}^2=\sigma_{u_0}^2+\sigma^2_\epsilon t\ $. Otherwise, the recurrence is known to have a solution of the form $\ \sigma_{u_t}^2=A+B\rho^{2t}\ $. Substituting this expression for $\ \sigma_{u_t}^2\ $ into the recurrence gives $\ A=\frac{\sigma_\epsilon^2}{1-\rho^2}\ $, and the initial condition at $\ t=0\ $ gives $\ B=\sigma_{u_0}^2-A\ $. In general, therefore, $$ \sigma_{u_t}^2=\cases{\sigma_0^2+\sigma^2_\epsilon t&if $\ |\rho|=1$\\\frac{\sigma_\epsilon^2}{1-\rho^2}+\left(\sigma_{u_0}^2-\frac{\sigma_\epsilon^2}{1-\rho^2}\right)\rho^{2t}&otherwise.} $$