uniformly convergent sequence of bounded functions is uniformly bounded

Can you explain in words the following? I'm confused about the last step in showing that $f_n$ is uniformly bounded. Why do we take the maximum over the previous $N$ functions, and how can we think of this as a bound for $n\to\infty$?

By uniform convergence of $\{f_n\}$ to $f$, we have $$||f_n - f||_{sup} < \epsilon, \quad \forall n \geq N$$

Assuming each $f_n$ is bounded by $M_n$, we can bound $f$ by

$$||f||_{sup} \leq ||f_N - f||_{sup} + ||f_N||_{sup} < \epsilon + M_N$$

Next, we bound $f_n$ by

$$||f_n||_{sup} \leq ||f_N - f||_{sup} + ||f||_{sup} < \epsilon + (\epsilon + M_N)$$

Finally, choose $M = \max\{M_1,\ldots,M_{n-1},M_N + 2\epsilon\}$ so that $f_n$ is uniformly bounded for all $n$.

The inequality $\|f_n\|\leqslant2\epsilon+M_N$ holds when $n\geqslant N$. If $n<N$, you know that $\|f_n\|\leqslant M_n$. Since $M\geqslant M_1\geqslant\|f_1\|$, $M\geqslant M_2\geqslant\|f_2\|$, ..., $M\geqslant M_{n-1}\geqslant\|f_{n-1}\|$ and $M\geqslant M_N+2\epsilon\geqslant\|f_n\|$, whenever $n\geqslant N$, it follows that $M$ uniformly bounds the whole sequence.