First denote by $\mathcal F(E)$ the set of all Fredholm operators mapping from $E \to E$; the index $\operatorname{ind} := \operatorname{dim}(\operatorname{ker}T) - \operatorname{dim}(E / T(E)$. We know that $\operatorname{ind}: \mathcal F(E) \to \mathbb Z $ is continuous and thus the sets $\mathcal F_k(E) := \{T \in \mathcal F(E): \operatorname{ind} T = k\}$ are open and mutually disjoint $\forall k \in \mathbb Z$.
I would argue as follows: Let $\lambda \neq 0 \in \sigma(K)$. First consider $I - {1 \over \lambda}K$ which has the same range and kernel as $\lambda I - K$. Then, $\{t \in [0,1]: \operatorname{ind}(I - t{1 \over \lambda}K)\}$ is path-connected and hence connected. But we know that $\operatorname{ind}$ is constant on each connected component and hence $\operatorname{ind}(I - {1 \over \lambda}K) = \operatorname{ind}(I) = 0$. Thus $\lambda I - K$ is injective if and only if it's surjective ($\star$).
By assumption $\lambda I - K$ is not bijective (otherwise by the Open Mapping Theorem, its inverse would be in $L(E)$). But then by ($\star$) it's not injective and thus $\operatorname{ker}(\lambda I - K) \setminus \{0\} \neq \varnothing$.
Anything to add?
The proof is correct, however I would note that $$ \left\{t \in [0,1]: \text{ind} \left(I−t\frac{1}{λ}K \right)\right\} $$ is not standard notation for what you mean (which I assume is the straight line from $I$ to $(I-\frac{1}{\lambda}T)$); what you should write is $$ \left\{ \text{ind} \left(I−t\frac{1}{λ}K \right) : t \in [0,1] \right\} .$$ This is standard notation as far as I know but if there is any reference you have of it being different I would be interested in seeing it.