It is more straightforward to solve this problem if we impose the condition that $s \leq t$ and calculate the problem using
$$
W_t = W_s + (W_t - W_s).
$$
Then,
$$
\text{E}\left[ W_s W_t \right] = \text{E} \left[ W_s \left( W_s + (W_t - W_s) \right) \right].
$$
Using the linearity of expectation, we have
$$
\text{E}\left[ W_s W_t \right] = \text{E} \left[ W_s^2 \right] + \text{E} \left[W_s (W_t - W_s) \right].
$$
Now, we use the fact that for two independent random variables
$$
\text{E}[AB] = \text{E}[A] + \text{E}[B]
$$
and the property of Brownian motion $\text{E}\left[ W_s^2 \right] = s$. So, we get
$$
\text{E}\left[ W_s W_t \right] = s + \text{E} \left[ W_s \right] \text{E} \left[W_t - W_s\right].
$$
We can see that the second term is zero as
$$
\text{E}\left[ W_s \right] = 0.
$$
Therefore,
$$
\text{E}\left[W_s W_t \right] = s.
$$
Recall that we imposed that $s \leq t$ initially. Therefore the general result is
$$
\text{E} \left[ W_s W_t \right] = \min(s, t).
$$