It is easy. For any matrix $||B||_{\infty}=\max_i\sum_j|b_{i,j}|\geq \max_i|b_{i,i}|$. Here $B=A^{-1}$ and $b_{i,i}=1/a_{i,i}$; then $||A^{-1}||_{\infty}\geq 1/\min(|a_{i,i}|)$.
The statement is false. Consider the max norm, like here. So the counterexample is:
$A := \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ is invertible and $B := \begin{pmatrix} 1/4 & 1/2 \\ 1/2 & 1 \end{pmatrix}$ is singular.
$$\|A^{-1}\| \cdot \|A-B\| = 1 \cdot \left\|\begin{pmatrix} 3/4 & -1/2 \\ -1/2 & 0 \end{pmatrix} \right\| = \frac{3}{4} < 1$$
The problem is indeed the absence of sub-multiplicative property. But if a matrix norm have this property then, the statement become true by the following lemma.
Lemma: Let $\|\cdot\|$ be a matrix norm on $\mathbb{R}^{n \times n}$. If $\|\cdot\|$ is sub-multiplicative, then there is a vector norm $\|\cdot\|_*$ on $\mathbb{R}^n$ such that both norms are compatible.
Proof: Fix $\boldsymbol{0} \neq \boldsymbol{y} \in \mathbb{R}^n$. Define $\|\cdot\|_* : \mathbb{R}^n \to [0,\infty)$ such that $\|\boldsymbol{x}\|_* := \|\boldsymbol{x}\boldsymbol{y}^T\|$ for every $\boldsymbol{x} \in \mathbb{R}^n$.
It's easy to check that $\|\cdot\|_*$ is a well defined vector norm because $\boldsymbol{y} \neq \boldsymbol{0}$ and the propierties of $\|\cdot\|$ for being, by hyphothesis, a sub-multiplicative matrix norm. So let's just check compatibility.
Let $A \in \mathbb{R}^{n \times n}$ and $\boldsymbol{x} \in \mathbb{R}^n$. Then by sub-multiplicative property of $\|\cdot\|$ we have
$$\|A\boldsymbol{x}\|_* = \|(A\boldsymbol{x})\boldsymbol{y}^T\| = \|A(\boldsymbol{x}\boldsymbol{y}^T)\| \leq \|A\| \cdot \|\boldsymbol{x}\boldsymbol{y}^T\| = \|A\| \cdot \|\boldsymbol{x}\|_*$$
Hence $\|\cdot\|$ is compatible with $\|\cdot\|_*$. $\blacksquare$
Then we are done since in the edited part of my post I had already demonstrated the result when we have sub-multiplicative and compatibility properties.
Best Answer
I mistakenly thought $\|\cdot\|$ was the $\ell^2$-norm. See below the line for the general situation.
Hint: $$\|Ax\|^2=\left\|\begin{bmatrix}\lambda_1 x_1 \\ \vdots \\ \lambda_n x_n\end{bmatrix}\right\|^2 = \lambda_1^2 x_1^2 + \cdots + \lambda_n^2 x_n^2 \le (\max_i \lambda_i^2) (x_1^2 + \cdots + x_n^2) = (\max_i \lambda_i^2) \|x\|^2.$$ By looking at the definition of $\|A\|$, can you now compute $\|A\|$? Computing $\|A^{-1}\|$ is similar, since it is also a diagonal matrix.
General situation:
Proof of Claim 1: Let $\|\cdot \|$ be a [submultiplicative] matrix norm. Let $x$ be a $\lambda_i$-eigenvector, and let $X$ be the $n \times n$ matrix whose columns are all $x$. Then $$|\lambda_i| \|X\| = \|\lambda_i X\| = \|A X\| \le \|A\| \|X\|.$$