Why does space(n) <= time(n) imply that TIME(f(n)) is a subset of SPACE(f(n))?
A machine which runs in time $f(n)$ also runs in space $f(n)$, that is, the space used by a machine is always at most the time it uses. In symbols, $\mathrm{space}_M(n) \leq \mathrm{time}_M(n)$.

A machine is in $\mathsf{TIME}(f(n))$ is it runs in time $O(f(n))$. Any such machine also uses space $O(f(n))$ (since the space used by a machine is at most the time it uses), and so it is in $\mathsf{SPACE}(f(n))$. In other words, $\mathsf{TIME}(f(n)) \subseteq \mathsf{SPACE}(f(n))$. This implies that
$$
\mathsf{P} = \bigcup_{k=0}^\infty \mathsf{TIME}(n^k) \subseteq \bigcup_{k=0}^\infty \mathsf{SPACE}(n^k) = \mathsf{PSPACE}.
$$
A similar calculation shows that $\mathsf{EXPTIME} \subseteq \mathsf{EXPSPACE}$.

For the remaining relation, we have to use a different fact about time and space: a machine which uses space $f(n)$ runs in time $O(2^{O(f(n))})$. This is because the number of configurations in a (one-tape) machine which uses space $f(n)$ is $|Q| f(n) |\Sigma|^{f(n)}$, where $|Q|$ is the number of states and $|\Sigma|$ is the size of the tape alphabet. You can check that this implies the remaining containment $\mathsf{PSPACE} \subseteq \mathsf{EXPTIME}$.