[proofplan]
We use the definition that a basis of a matroid is a maximal independent subset of the ground set. If $B$ is maximal independent, then every element outside $B$ is forced into the closure of $B$, because adding it creates dependence. Conversely, if $B$ is independent and spans $E$, then every element outside $B$ is already in $\operatorname{cl}_M(B)$, so adjoining it destroys independence; this makes $B$ maximal independent.
[/proofplan]
[step:Show that a maximal independent set spans the ground set]
Assume that $B$ is a basis of $M$. By definition of basis, $B \in \mathcal{I}$ and $B$ is maximal among the independent subsets of $E$ under inclusion.
We prove that $\operatorname{cl}_M(B) = E$. Since closure is extensive, $B \subset \operatorname{cl}_M(B) \subset E$. Let $e \in E \setminus B$. Because $B$ is maximal independent, the set $B \cup \{e\}$ is not independent. Since $B$ is independent, the closure criterion for matroids gives
\begin{align*}
e \in \operatorname{cl}_M(B).
\end{align*}
Thus every element of $E \setminus B$ lies in $\operatorname{cl}_M(B)$, and every element of $B$ lies in $\operatorname{cl}_M(B)$ by extensivity. Hence $\operatorname{cl}_M(B) = E$. Therefore $B$ is independent and spanning.
[/step]
[step:Show that an independent spanning set is maximal independent]
Assume that $B \in \mathcal{I}$ and $\operatorname{cl}_M(B) = E$. We prove that $B$ is maximal independent.
Let $e \in E \setminus B$. Since $\operatorname{cl}_M(B) = E$, we have $e \in \operatorname{cl}_M(B)$. Because $B$ is independent, the closure criterion for matroids implies that $B \cup \{e\}$ is dependent. Therefore no single element of $E \setminus B$ can be added to $B$ while preserving independence.
Now let $S \subset E$ be any subset satisfying $B \subsetneq S$. Choose $e \in S \setminus B$. Then
\begin{align*}
B \cup \{e\} \subset S.
\end{align*}
Since $B \cup \{e\}$ is dependent, the hereditary property of matroid independence implies that $S$ is not independent. Hence no proper superset of $B$ is independent, so $B$ is maximal independent. Therefore $B$ is a basis of $M$.
[guided]
Assume that $B \in \mathcal{I}$ and $\operatorname{cl}_M(B) = E$. To prove that $B$ is a basis, we must prove maximality among independent subsets of $E$: every proper superset of $B$ must fail to be independent.
First take a single element $e \in E \setminus B$. Since $B$ spans $E$, the equality $\operatorname{cl}_M(B) = E$ gives
\begin{align*}
e \in \operatorname{cl}_M(B).
\end{align*}
For an independent set $B$, membership of $e$ in the closure of $B$ means exactly that adjoining $e$ creates dependence. Thus
\begin{align*}
B \cup \{e\} \notin \mathcal{I}.
\end{align*}
So no element outside $B$ can be added one at a time while preserving independence.
It remains to rule out larger proper supersets, not just one-element extensions. Let $S \subset E$ satisfy $B \subsetneq S$. Since the inclusion is proper, there exists an element $e \in S \setminus B$. Then $B \cup \{e\}$ is a subset of $S$:
\begin{align*}
B \cup \{e\} \subset S.
\end{align*}
We already proved that $B \cup \{e\}$ is dependent. If $S$ were independent, then every subset of $S$ would be independent by the hereditary axiom for matroids, contradicting the dependence of $B \cup \{e\}$. Therefore $S$ is dependent.
Since every proper superset $S$ of $B$ is dependent, the independent set $B$ is maximal independent. By definition, a maximal independent subset of the ground set is a basis of $M$. Hence $B$ is a basis.
[/guided]
[/step]
