Let $Y=(\mathbb{R}^{\mathbb{N}},\|\cdot\|_1)$ where $\|x\|_1=\sum_{n=1}^\infty|x_n|$, and $X=(\mathbb{R}^{\mathbb{N}},\|\cdot\|_X)$ where $\|x\|_X=\|x\|_1+\sup_{n\in\mathbb{N}}n|x_n|$. Then $\|\cdot\|_X$ is a norm since $\|ax\|_X=|a|\|x\|_X$ for a scalar $a$, $\|x\|_X=0$ implies $x=0$, and \begin{align} \|x_1+x_2\|_X&=\|x_1+x_2\|_1+\sup_{n\in\mathbb{N}}n|x_{1n}+x_{2n}| \\ &\leq\|x_1\|_1+\|x_2\|_1+\sup_{n\in\mathbb{N}}n|x_{1n}|+\sup_{n\in\mathbb{N}}n|x_{2n}| \\ &=\|x_1\|_X+\|x_2\|_X. \end{align} Note that $\|x\|_1<\infty$ if and only if $\|x\|_X<\infty$ and that $\|x\|_1\leq\|x\|_X$. Let $T:X\to Y$ be identity, which is continuous, linear, and bijective. Then, $T^{-1}$ exists and is linear, but is not continuous.
Which assumption of the inverse operator theorem is violated in this example?
The map is not onto. The condition on $X$ is stricter than $Y$. Note the sequence given by $x_n=\frac{1}{k^2}$ if $n=k!$ and $x_n=0$ otherwise is in $Y$ but not in $X$.