We are going to need the following elementary lemma:
Lemma: Let $X$ be a normed space and $M$ a closed subspace. Then the quotient map $q:X\to X/M$ is an open map.
Proof of lemma without using the open mapping theorem: Let $U\subset X$ be an open set. To show that $q(U)$ is open, let $x\in U$; by $U$'s openness find $r>0$ such that $B(x,r)\subset U$. Now let $x'+M\in X/M$ be such that $\|(x+M)-(x'+M)\|<r$. Then $\|(x-x')+M\|<r$, i.e. $\inf_{m\in M}\|x-x'+m\|<r$. Since this infimum is strictly smaller than $r$, we obtain some $m\in M$ such that $\|x-x'+m\|<r$. But this shows that $x'-m\in B(x,r)\subset U$, so $x'-m\in U$ and thus $q(x'-m)\in q(U)$ and note that $q(x'-m)=q(x')=x'+M$. We have just shown that $B(x+M,r)\subset q(U)$, proving that $q(U)$ is open.
Assume now that we know the inverse mapping theorem, i.e. if $T:X\to Y$ is a bounded linear operator between Banach spaces that is a bijection, then it's inverse $T^{-1}:Y\to X$ (which is obviously linear) is also bounded.
Let $T:X\to Y$ be a bounded surjective map between Banach spaces. We shall show that $T$ has to be an open map. To do that, let $M=\ker(T)$. This is a closed subspace of $X$ so we can consider the quotient space $X/M$. Note that $T$ induces a bounded linear operator $\bar{T}:X/M\to Y$ by $\bar{T}(x+M)=Tx$. Let's check this: Indeed $\bar{T}$ is well-defined: if $x+M=x'+M$, then $x-x'\in M=\ker(T)$ so $Tx=Tx'$. Linearity is obvious. For boundedness, note that for any $m\in M$ we have $$\|\bar{T}(x+M)\|=\|Tx\|=\|T(x+m)\|\le\|T\|\cdot\|x+m\|$$
and since this is true for any $m\in M$ we can take the infimum over $m$ in the above inequality to conclude that
$$\|\bar{T}(x+M)\|\le\|T\|\cdot\inf_{m\in M}\|x+m\|=\|T\|\cdot\|x+M\|_{X/M}$$
It is also obvious that $T$ is bijective: if $T(x+M)=0$ then $Tx=0$ so $x\in M$, i.e. $x+M=M=0_{X/M}$, showing injectivity. If $y\in Y$ then find $x\in X$ with $Tx=y$, by $T$'s assumed surjectivity; thus $y=\bar{T}(x+M)$, proving surjectivity of $\bar{T}$. Apply the inverse mapping theorem (note that $X/M$ is a Banach space itself, since $M$ is a closed subspace of a Banach space!) to $\bar{T}$ to obtain a bounded linear operator $S:Y\to X/M$ with the property that $S\bar{T}=\text{id}_{X/M}$ and $\bar{T}S=\text{id}_Y$.
Now let $U\subset X$ be an open set. Then $U+M$ is open in $X/M$, by our lemma. Now since $S$ is continuous, we have that $S^{-1}(U+M)$ is open in $Y$, by definition of continuity. But since $S^{-1}=\bar{T}$, we have that $S^{-1}(U+M)=\bar{T}(U+M)=T(U)$: let's prove the last equality; if $x\in U$, then $Tx=\bar{T}(x+M)\in\bar{T}(U+M)$, showing that $T(U)\subset\bar{T}(U+M)$. For the other inclusion, if $x\in X$ with $x+M\in U+M$, then find $x'\in U$ such that $x+M=x'+M$. But then $\bar{T}(x+M)=\bar{T}(x'+M)=Tx'\in T(U)$.
Since $\text{open}=S^{-1}(U+M)=T(U)$, we are done.
Best Answer
Since you have worked out that $T$ must be surjective we know that $T$ must be a bijection and so there exists a linear map $T^{-1}:Y \to X$ that is inverse to $T$. It remains to see that $T^{-1}$ is continuous.
By the topological characterisation of continuity, we need to show that if $O$ is open in $X$ then $T(O) = (T^{-1})^{-1}(O)$ is open in $Y$. This is immediate since $T$ is an open map. Hence $T$ is continuous which is what we wanted to prove.