Bounded operator has closed image if and only if there exists positive a constant $c$ such that $c \| x \| \leq \| Tx \|$. In my notes I have the following theorem

If $T:X \to Y$ is a bounded operator where $X,Y$ are complex Hilbert spaces. Prove that $Im (T)$ is a closed set if and only if there exists positive constant $c$ such that $c \| x \| \leq \| Tx \|$ for $x \in (\ker T )^{\perp }$.

The proof of '$\impliedby$' I understand, but the proof my notes give for '$\implies$' is very brief and I don't understand.
Here's the proof.
Suppose we have the operator $T_1 : (\ker T)^{\perp } \to Im(T)$ such that $T_1x =Tx$, which is bijection and continuous. Then by the open mapping theorem, the desired inequality follows.
It's not obvious to me why the inequality follows. Also I can't see why $T_1$ is a bijection.
My understanding so far is this.
$T_1$ is continuous since $T$ is bounded.
$Im(T)$ is closed so $\overline{Im(T)} =Im(T)$, if it was true that $Im(T)$ is dense in $Y$ , we would get that $T_1$ is surjective, but I can't see if that's the case.
I don't understand why is $T_1$ injection.
Now for the inequality, since $T$ is continuous and bijection then $T^{-1}$ exists and is bounded by say $\frac{1}{c}$, so $$\| x \| =\| T^{-1} T x \| =\| T^{-1} \| \| Tx \| \leq \frac{1}{c} \| Tx \|.$$
So $c \| x \| \leq \| Tx \|$.
I also don't see where the open mapping theorem fits in.
Can you explain the proof?
 A: Consider the map between Hilbert spaces ( the closedness of the image ensure completeness) $\operatorname{ker}(T)^{\perp} \to \operatorname{Im}(T)$. It is bijective and continuous, so it is bicontinuous (Banach open  mapping theorem).  You have to check that it is bijective ( use $X= \operatorname{Ker}(T) \oplus \operatorname{Ker}(T)^{\perp}$).
A: $x \in Ker (T)^{\perp}, T_1x=0$ implies $Tx=0$. But then $x \in Ker (T) \cap Ker (T)^{\perp}= \{0\}$ so $x=0$. Hence, $T_1$ is injective.
If $y \in Im (T)$ then there exists $x \in X$ such that $y=Tx$. Since $X=Ker (T)+Ker (T)^{\perp} $ we can write $x=x_1+x_2$ with $x_1 \in Ker (T)$ and $x_2 \in Ker (T)^{\perp}$.  Now $y=Tx=Tx_1+Tx_2=0+Tx_2=T_1x_2$ which proves that $y \in Im (T_1)$. Hence, $T_1$ is surjective.
To prove that $Im (T)$ is closed take a sequence $(y_n)$ in it converging to some $y \in Y$.  Let $y_n=Tx_n$ and note that $c\|x_n-x_m\| \leq \|y_n-y_m\| \to 0$. Hence, $(x_n)$ is Cauchy sequence. Let $x_n \to x$. Then $y=Tx$ by continuity, completing the proof.
Open Mapping Theorem is needed to say that a bijective continuous linear map between Banach spaces has a continuous inverse.
