Proof : $S$ a subspace of a vector space is generated by $n$ vectors, any $m$ vectors in $S$ , where $m > n$ are linearly dependent. Constraints :

*

*Basis and dimensions are not introduced yet

*Matrix and system of equations can't be used

I am aware that this might possibly be a duplicate of the same question here, but I wanted to take this question further and see if any of you have come across an alternative proof than the one mentioned in the first answer. Btw, I am new to this platform, so in case I am making a mistake by posting this, please let me know.
 A: The following proof may resemble the second answer of the linked post, but I do not fully understand the argument there.
Clearly, the assertion in the title follows from Lemma 1 below (if $v_1,\cdots,v_n$ are linearly dependent, then the space $S$ can be spanned by fewer than $n$ vectors).
Lemma 1. Let $v_1,\cdots,v_n$ be independent and $$S=<v_1,\cdots,v_n>.$$ Then any $n+1$ vectors in $S$ are linearly dependent.
Lemma 2. Assume Lemma 1 holds for $n=k-1$. Let $$S=<v_1,\cdots,v_k>,~{\rm and~}w_1,\cdots,w_k\in S,$$ where $v_1,\cdots,v_k$ (resp. $w_1,\cdots,w_k$) are linearly independent. Then $$S=<w_1,\cdots,w_k>.$$
Proof. Given the assumption, it suffices to show that $$S=<v_1,\cdots,v_k>=<w_i,v_2,\cdots, v_k>\quad (1)$$ for some $1\leq i\leq k$ and $w_i,v_2,\cdots,v_k$ are linearly independent. Repeating the argument, one can achieve the result by replacing all $v_i$’s by $w_i$’s up to ordering.
To prove (1), one uses the assumption that Lemma 1 holds for $n=k-1$. It follows that there exists $i$ such that $$w_i\notin ~<v_2,\cdots,v_k>,$$ since $w_1,\cdots,w_k$ are linearly independent. Now $$w_i=c_1v_1+c_2v_2+\cdots+c_kv_k~{\rm with~}c_1\neq 0,$$ hence $$v_1\in ~<w_i,v_2,\cdots,v_k>~\supseteq ~<v_1,\cdots,v_k>$$
$$\Rightarrow ~<w_i,v_2,\cdots,v_k>=<v_1,\cdots,v_k>,$$ since the other inclusion is trivial. The independence of $w_i,v_2,\cdots,v_k$ can be verified by definition. $\Box$
Proof of Lemma 1. One proves it by induction on $n$, the case $n=1$ being clear. Assume the case $n=k-1\geq 1$ is true. One needs to show that case for $n=k$: Namely if $S=<v_1,\cdots,v_k>$, and $v_1,\cdots,v_k$ are linearly independent, then any $k+1$ vectors $w_1,\cdots,w_k,w_{k+1}$ in $S$ are linearly dependent.
If $w_1,\cdots,w_k$ are linearly dependent, the result is clear. Hence one may assume that $w_1,\cdots,w_k$ are linearly independent. But then by Lemma 2, one has $$S=<v_1,\cdots,v_k>=<w_1,\cdots,w_k>.$$ Since $w_{k+1}\in S=<w_1,\cdots,w_k>,$ one has $$w_{k+1}=\sum_{i=1}^k c_iw_i$$ $$\Rightarrow w_1,\cdots,w_{k+1}$$ are linearly dependent. QED
A: For comparison with my previous writing, the following is a slightly different write up.
The result of the OP’s assertion follows from the following lemma. One makes sure not to use matrices, linear equations, bases, or dimension.
Lemma. Let $v_1,\cdots,v_n$ be independent and $$S=<v_1,\cdots,v_n>.$$ Then any $n+1$ vectors in $S$ are linearly dependent.
Proof. One proceeds by induction on $n$, the case $n=1$ being clear.
(IH) Assume that the result is true for $n=k-1\geq 1,$ i.e. if $v_1,\cdots,v_{k-1}$ are linearly independent and $T=<v_1,\cdots,v_{k-1}>.$ Then any $k$ vectors in $T$ are linearly dependent.
For $n=k$, let $v_1,\cdots,v_k$ be linearly independent and $S=<v_1,\cdots,v_k>.$ Let $w_1,\cdots,w_{k+1}\in S.$ One needs to show that $w_1,\cdots,w_{k+1}$ are linearly dependent.
Case 1. $w_1,\cdots,w_k$ are linearly dependent.
Clearly $w_1,\cdots,w_{k+1}$ must be linearly dependent in this case.
Case 2. $w_1,\cdots,w_k$ are linearly independent.
Claim: $S=<v_1,\cdots,v_k>=<w_1,\cdots,w_k>.$
Proof of Claim: It suffices to show that $$S=<v_1,\cdots,v_k>=<w_i,v_2,\cdots,v_k>$$ for some $1\leq i \leq k,$ and $w_i,v_2,\cdots,v_k$ are linearly independent. ......(1)
(Repeating the argument, one can achieve the Claim by replacing all $v_i$’s by $w_i$’s up to reordering.)
To prove (1), note that by induction hypothesis (IH) applied to $T=<v_2,\cdots,v_k>,$ one sees that not all $w_j$’s are in $T$, since $w_1,\cdots,w_k$ are linearly independent. It follows that there exists $i,1\leq i\leq k$ such that $w_i\notin T=<v_2,\cdots,v_k>.$ Now $$w_i=c_1v_1+c_2v_2+\cdots+c_kv_k~{\rm with~}c_1\neq 0,$$ hence $$v_1\in ~<w_i,v_2,\cdots,v_k>~\supseteq~<v_1,\cdots,v_k>$$
$$\Rightarrow <w_i,v_2,\cdots,v_k>=<v_1,\cdots,v_k>,$$ since the other inclusion is trivial. The independence of $w_i,v_2,\cdots,v_k$ can be verified by definition. This concludes the proof of (1), hence the Claim.
Now to finish Case 2 and show that $w_1,\cdots,w_k,w_{k+1}$ are linearly dependent, note by the Claim that $w_{k+1}\in S=<w_1,\cdots,w_k>,$ hence $$w_{k+1}=\sum_{i =1}^k c_iw_i$$ $$\Rightarrow w_1,\cdots,w_{k+1}~{\rm are ~linearly ~dependent.}$$
This concludes the proof for Case 2, and the Lemma. QED
