Here is a way to show this when you consider the field of real numbers and this for all norms.
Consider $E$ a finite dimensional vector space over $\mathbb{R}$ and a basis $(e_1, ..., e_n)$ of $E$. It means that for all $x\in E$ you have $x = \sum_{i=1}^{n} x_{i}e_i $ where the $x_i$ are real numbers.
Now we need to endow E with a norm, first recall that all norms are equivalents in a finite dimensional normed space it will be usefull at some points.
The norm we will consider is $\lVert x\rVert_{\infty} = max_{1\leq i\leq n}\lvert x_i\rvert $. Thus $(E,\lVert\rVert_{\infty})$ is a finite dimensional normed space.
Now consider a Cauchy sequence in $E$, namely $(x^{p})_{p\in\mathbb{N}}$, it means that
$\forall\epsilon>0,\exists N\in\mathbb{N} : n,m\geq N\implies\lvert x_i^{n} -x_i^{m}\rvert\leq\lVert x^{n} - x^{m}\rVert_{\infty} = max_{1\leq i \leq n}\lvert x_{i}^{n} - x_{i}^{m}\rvert\leq\epsilon$
However for $i\in\{1,...,n\}$ we have that $(x_i^{p})$ is a sequence of real numbers and we have shown that it is Cauchy !
Thus for all $i\in\{1,..n\}$ there exists $x_i\in\mathbb{R}$ such that $\forall\epsilon>0, \exists N_{i}\in\mathbb{N} : p\geq N_i\implies \lvert x_i^{p} -x_i\rvert\leq\epsilon$ (since $\mathbb{R}$ is complete)
Now, a candidate we should consider as a limit of $(x^p)_{p\in\mathbb{N}}$ is the vector whose coordinate corresponds to the $x_i$'s that is $x = \sum_{i=1}^{n}x_ie_i$.
Fix $\epsilon>0$ and $N = max_{1\leq i\leq n}(N_i)$, then we have that
$p\geq N\implies max_{1\leq i\leq n}\lvert x_i^p - x_i\rvert = \lVert x^p - x\rVert_{\infty}\leq\epsilon$ which shows the convergence of $(x^p)_{p\in\mathbb{N}}$ to $x\in E$.
Since $E$ is a finite dimensional normed space, all norms on this space are equivalents which means that the convergence is not altered by a change of norm (this is already explained in the comments I think)
I hope this is clear !
The relation $\dim(\mathrm{coker}(T))<\infty\iff\dim(\ker(T^*))<\infty$ remains true in arbitrary Banach spaces:
If $T$ has finite dimensional co-kernel then $\overline{\mathrm{im}(T)}$ admits a finite dimensional complement, choose one such complement and call it $V$. Note that $T^*(f)=0$ iff $f(Tv)=0$ for all $v\in X$, ie iff $f\lvert_{\overline{\mathrm{im}(T)}}=0$. As such $f$ is uniquely determined by its values on $V$, ie the map $\ker(T^*)\to V^*$ given by $f\mapsto f\lvert_V$ is injective. But $V^*$ is finite dimensional, so $\ker(T^*)$ also is finite dimensional.
On the other hand if the co-kernel is not finite dimensional then $\overline{\mathrm{im}(T)}$ admits infinite dimensional (not necessarily closed though) complements. Now you can check for any $V$ finite dimensional and linearly independent to $\overline{\mathrm{im}(T)}$ that for any $f\in V^*$ the map $V\oplus \mathrm{im}(T)\to\Bbb C, (v,x)\mapsto f(v)$ is continuous. In particular it admits Hahn Banach extensions with domain all of $X$. But any such Hahn Banach extension is $0$ on $\overline{\mathrm{im}(T)}$ hence lies in $\ker(T^*)$. Since you can do this for all $f\in V^*$ where $V$ has arbitrary finite dimension you must find that $\ker(T^*)$ is infinite dimensional.
Best Answer
Make use of the following facts:
Use this complement to see that for $x\in A$:
\begin{align} \|x\|_A &=\|Px+(1-P)x\|_A≤ \|Px\|_A+\|(1-P)x\|_A ≤ k\,\|Px\|+\|(1-P)x\|_A\\ &≤k\,\|Px+(1-P)x-(1-P)x\|+\|(1-P)x\|_A≤ k\,\|x\|+k\,\|(1-P)x\|+\|(1-P)x\|_A\\ & ≤ k\,\|x\|+(1+kC)\|(1-P)x\|_A. \end{align} Now $(1-P)x$ is in $Y$. The map $S:Y\to\mathrm{im}(S)$ is continuous, injective, surjective and has codomain a Banach space (because we know that $\mathrm{im}(S)$ is closed). Hence it is invertible. This means that there is a constant $K2$ with $\|(1-P)x\|_A ≤ K\, \|S(1-P)x\|_B$. Note that $S(1-P)=S$, hence $\|(1-P)x\|≤K\,\|Sx\|_B$. Combine both inequalities to get: $$\|x\|_A ≤ k\,\|x\|+(1+kC)K\,\|Sx\|_B.$$