Every linear operator $T:X \to Y$ on a finite-dimensional normed space is bounded
Solution 1:
Your proof is both correct and rigorous. The only change I would suggest is that, instead of using Cauchy-Schwarz, you can take $M=\max\{\|Te_k\|:\ k=1,\ldots,n\}$ and then $$ \sum_{k=1}^{n} |a_k| \| T(e_k) \|_Y \leq M\,\sum_{k=1}^n|a_k|, $$ and you can use that norm to compare with the original.