Difficulty in understanding the proof of the lemma : "Matroids exhibit the optimal-substructure property"
The adjective &quot;maximum-weight&quot; should not appear in item (1) in that proof of the lemma. This is a minor bug of that famous book.
To be fully clear, item (1) should have been the following.

If $A$ is any independent subset of $M$ containing $x$, then $A' = A - \{x\}$ is an independent subset of $M'$.

With item (1) corrected, item (4) follows naturally from item (1), (2) and (3). Here is more detail.

&quot;A maximum-weight solution in $M$ containing $x$ yields a maximum-weight solution in $M'$.&quot;
Note that &quot;solution&quot; is just a shorthand for &quot;independent set&quot;. Let us prove the proposition above.
Suppose $A$ is a maximum-weight solution in $M$. Then $A$ yields $A'=A-\{x\}$, which is a solution in $M'$ according to item (1). (The previous version of item (1) works as well.)
Given any solution $B'$ in $M'$, let $B=B'\cup\{x\}$, which is a solution in $M$ according to item (2).
Item (3) tells us $w(A)=w(A')+w(x),$ and $w(B)=w(B')+w(x).$ Since $A$ has maximum weight in $M$, we have $w(A)\ge w(B)$, i.e.,
$$w(A')+w(x)\ge w(B')+w(x).$$
Cancelling $w(x)$ from both sides, we obtain
$$w(A')\ge w(B'),$$
which says $A'$ is a maximum-weight solution in $M'$. $\checkmark$

A maximum-weight solution in $M'$ yields a maximum-weight solution in $M$ containing $x$.
The other direction, as stated above, can be proved similarly. Here is the proof.
Suppose $B'$ is a maximum-weight solution in $M'$. Then $B'$ yields $B=B'\cup\{x\}$, which is a solution in $M$ according to item (2).
Given any solution $A$ in $M$, let $A'=A-\{x\}$, which is a solution in $M'$ according to (the corrected version of) item (1).
Since $B'$ has maximum weight in $M'$, we have $w(B')\ge w(A')$. Adding $w(x)$ to both sides, we obtain,
$$w(B')+w(x)\ge w(A')+w(x).$$
Item (3) tells us $w(A)=w(A')+w(x),$ and $w(B)=w(B')+w(x).$  So the inequality above is the same as
$$w(B)\ge w(A),$$
which says $B$ is a maximum-weight solution in $M$. $\checkmark$