Let $V$ be an $F$-vector space, $I$ an index set, and $j ∈ I$ such that $(v_i)_{i ∈ I \setminus \{j\}}$ is a linearly independent set of vectors in $V$.
Required to prove: $(v_i)_{i ∈ I}$ is linearly independent iff $v_j \notin L(\{v_i ∈ V \mid i ∈ I \setminus \{j\}\})$.
I have already proved '⇒', so let's assume that $v_j \notin L(\{v_i ∈ V \mid i ∈ I \setminus \{j\}\})$. We want to show that for any finite subset $T ⊆ I$ and, for $\lambda_i ∈ F$ such that $\sum_{i ∈ T} \lambda_iv_i = 0$, we have $\{\lambda_i ∈ F \mid i ∈ T\} = \{0\}.$
So let $T ⊆ I$ be finite and $\{\lambda_i ∈ F\mid i ∈ T\}$ such that $\sum_{i ∈ T} \lambda_iv_i = 0$. If $j \notin T$ then $T ⊆ I \setminus \{j\}$, so the result is immediate from the assumption that $(v_i)_{i ∈ I \setminus \{j\}}$ is linearly independent. We can therefore assume, w.l.o.g., that $j ∈ T$. Let us now define $T' := T \setminus \{j\} ⊆ I \setminus \{j\}$. Then by the assumtion that $v_j$ is not in the span of the other $v_i$, we have $\sum_{i ∈ T'} \lambda_iv_i ≠ v_j$, and by the assumption of the linear independence of $(v_i)_{i ∈ I \setminus \{j\}}$ we have the implication $\sum_{i ∈ T'}\lambda_iv_i = 0 \Rightarrow \{\lambda_i \in F \mid i ∈ T'\} = \{0\}$.
Note that we now have $$0 = \sum_{i ∈ T} \lambda_iv_i = \sum_{i ∈ T'} \lambda_iv_i + \lambda_jv_j. $$
I suppose we can again distinguish two cases, namely A) $\sum_{i ∈ T'} \lambda_iv_i= 0$ or B) $\sum_{i ∈ T'} \lambda_iv_i ≠ 0$.
In case A, the aforementioned implication proves that $\{\lambda_i ∈ F \mid I ∈ T'\} = \{0\}$, so it follows that $0 = \sum_{i ∈ T'}0 · v_i + \lambda_jv_j$, whence $\lambda_j = 0$ and the result follows.
In case B, however, I now arrive at $$0 = \sum_{i ∈ T} \lambda_iv_i = \sum_{i ∈ T'} \lambda_iv_i + \lambda_jv_j \notin \{\lambda_jv_j, (\lambda_j+1)v_j\}. $$
From this, we get $\lambda_j \notin \{0, -1\}$. But I need $\lambda_j$ to be $0$...!! So did I make the mistake in thinking that case B can even occur? Or am I making another false assumption?
You've almost got it. At the end of your proof, if $\lambda_j\ne 0$, then you can write $$v_j=\sum_{i\in T'}-\lambda_j^{-1}\lambda_i v_i$$ which contradicts the fact that $v_j$ is not in the span of the $v_i$. So you must have $\lambda_j=0$, and then the rest of the $\lambda_i=0$ by the assumption of linear independence.