Why is \(cov(x_t, x_{t+k}) = t \sigma^2\)?
First, note that that since the terms in the white noise series are independent,
\[
cov ( w_i, w_j ) =
\begin{cases}
\sigma^2, & \text{if } ~ i=j \\
0, & \text{otherwise}
\end{cases}
\]
Also, when random variables are independent, the covariance of a sum is the sum of the covariance.
Hence, \[\begin{align*}
cov(x_t, x_{t+k})
&= cov ( \sum_{i=1}^t w_i, \sum_{j=1}^{t+K} w_j ) \\
&= \sum_{i=j} cov ( w_i, w_j ) \\
&= \sum_{i=1}^t \sigma^2 \\
&= t \sigma^2
\end{align*}\]