Show that $T^{-1}:Y \to X$ exists and is bounded. 
Let $T$ be a bounded linear operator from a normed space $X$ onto a normed space $Y$. If there is a positive $b$ such that $$||Tx||\ge b||x||,$$ for all $x \in X$, show that $T^{-1}:Y \to X$ exists and is bounded.

My attempt:
Suppose $Tx =0$ then clearly $||Tx~||=0 \iff || x||=0$ and so $T$ is injective, that is $T^{-1}$ exists. We need now show that $T^{-1}$ is bounded.
This is where I am stuck. I have the following trail of thought, but I am not sure if this is correct: 
Since they say $T$ is a b.l.o. from $X$ onto $Y$, we can say that $T$ is surjective. Now, for every $y \in Y$ then there exists an $x=T^{-1}y \in X$ and so \begin{align}||T(T^{-1}y)|| &\ge b||T^{-1}y|| \\
\therefore ||y|| &\geq b||T^{-1}y|| \\
\therefore \frac{1}{b}||y|| &\ge ||T^{-1}y||,\end{align} for every $y \in Y$. That is, $T^{-1}$ is bounded.
Is this correct?

As I said in a comment, you should also check that $T^{-1}$ is linear but the rest seems fine. Here is the proof.
Let $y_1, y_2 \in Y$ and $\lambda, \mu \in \mathbb{C}$. Using the linearity of $T$ you've got
$$ \lambda y_1 + \mu y_2 = \lambda TT^{-1}(y_1)+ \mu TT^{-1}(y_2) \\= T\Big(\lambda T^{-1}(y_1) + \mu T^{-1}(y_2)\Big).$$
Then apply $T^{-1}$ on both sides :
$$ T^{-1}\big(\lambda y_1 + \mu y_2\big) = \lambda T^{-1}(y_1) + \mu T^{-1}(y_2).$$
