What's wrong with this proof of almost sure limits?

As pointed out by snar, the problem in your approach is that the obtained bound for the maximum depends on $n$, that is $Y_n\leqslant M_n^n$ where $M_n=\max_{1\leqslant k\leqslant n}X_k$. And it is a priori possible that $M_n$ converges to $1$ hence taking the limit may lead to an undetermined form. For example if $X_i=1-1/i$, then $M_n=1-1/n$ and $M_n^n\to 1/e$.

However, the previous configuration $X_i=1-1/i$ is actually almost surely impossible. The point is that for almost every $\omega$, an infinite amount of $X_n(\omega)$ will be smaller than $1/2$. This follows from the second Borel-Cantelli lemma applied to the sequence of independent events $A_n=\left\{X_n\leqslant 1/2\right\}$.

$\ln Y_n= \sum_{i=1}^n \ln X_i$, where $-\ln X_i$'s are i.i.d. $\exp(1)$ r.v.s. By the SLLN, $\ln Y_n^{1/n}\to -1$ a.s. and $Y_n^{1/n}\to e^{-1}$ a.s. Now use the fact that for each $k\ge 1$, $\limsup_{n\to\infty}Y_n\le e^{-k}$ a.s.