I was asked to prove the following Theorem.
Theorem. Given that $V$ is finite dimensional and $T\in\mathcal{L}(V)$ and $\lambda\in\mathbf{F}$. There exists a $\alpha\in\mathbf{F}$ such that $|\alpha-\lambda|<\frac{1}{1000}$ and $T-\alpha I$ is invertible.
The Following is my attempt any extra results that i make use of have been quoted below the proof. Is my proof correct?
Proof. Assume on the contrary that given any $\alpha\in\mathbf{F}$, $|\alpha-\lambda|<\frac{1}{1000}$ implies that $T-\alpha I$ is not invertible or equivalently $\alpha$ is an eigenvalue of $T$.
Let $n = \dim V$ and $I_{n+1} = \{1,2,3,...,n,n+1\}$. Consider now the function $\mathcal{V}:I_{n+1}\to \mathbf{F}$ defined by $$\mathcal{V}(x) = \lambda+\frac{1}{10^{3+x}}$$
with the above definition it is not difficult to see that $\forall x\in I_{n+1}\left(|\mathcal{V}(x)-\lambda|<\frac{1}{1000}\right)$.
Consequently we have a list of $n+1$ distinct eigenvalues namely $\mathcal{V}(1),\mathcal{V}(2),...,\mathcal{V}(n),\mathcal{V}(n+1)$ taking this together with theorem $5.10$ implies that the corresponding list of eigenvectors namely $v_1,v_2,...,v_n,v_{n+1}$ is linearly independent.
However we know that any list of linearly independent vectors in $V$ must have length $m\leq n = \dim V$ resulting in a contradiction.
$\blacksquare$
$5.10$ If $T\in\mathcal{L}(V)$ and $\lambda_1,\lambda_2,...,\lambda_n$ is a list of distinct eigenvalues of $T$ then the corresponding list of vectors $v_1,v_2,...,v_n$ is linearly independent.