Steinitz exchange lemma

Argue by induction on $k$, the number of $w$'s exchanged in so far. For $k=0$ the inequality $0\leq m$ is trivial, and no exchange is needed. Suppose inductively that $w_1,\dots,w_{k-1}$ have already been exchanged in for $k-1$ of the original $v_i$'s (relabel so these are $v_1,\dots,v_{k-1}$), so that $\{w_1,\dots,w_{k-1},v_k,\dots,v_m\}$ still spans $V$.

If $k-1$ already equals $m$, every $v_i$ has been used up, so $\{w_1,\dots,w_{k-1}\}$ alone spans $V$; but then $w_k\in V$ would be a linear combination of $w_1,\dots,w_{k-1}$, contradicting the linear independence of $\{w_1,\dots,w_k\}$. So this case cannot occur while $k\leq m$ is still to be shown for the current $k$ — precisely, it shows $k-1<m$, i.e. $k\leq m$, immediately.

Otherwise $k-1<m$, so at least one $v_j$ (among $v_k,\dots,v_m$) remains. Since $\{w_1,\dots,w_{k-1},v_k,\dots,v_m\}$ spans $V$, write $w_k$ as a combination $w_k=\lambda_1w_1+\cdots+\lambda_{k-1}w_{k-1}+\mu_1v_{j_1}+\cdots+\mu_{m-k+1}v_{j_{m-k+1}}$. If every coefficient $\mu_1,\dots$ on the remaining $v$'s were $0$, then $w_k$ would be a combination of $w_1,\dots,w_{k-1}$ alone, again contradicting independence — so some remaining $v_j$ has a nonzero coefficient.

Solve that equation for $v_j$ (dividing by its nonzero coefficient), expressing $v_j$ as a combination of $w_1,\dots,w_k$ and the other remaining $v$'s. Substituting this expression for $v_j$ wherever it appears shows that $\{w_1,\dots,w_k\}$ together with the remaining $v$'s (minus $v_j$) still spans $V$ — one more $v$ has been successfully exchanged for $w_k$, completing the inductive step and confirming $k\leq m$.