Passing from induction to $\infty$

What your argument shows is that, for any sequence $a_k$ in $X$, we always have $$\bigg\|\sum_{k=1}^na_k\bigg\|\leq\sum_{k=1}^n\|a_k\|$$ for any $n$, and therefore $$\lim_{n\to\infty}\bigg\|\sum_{k=1}^na_k\bigg\|\leq\lim_{n\to\infty}\sum_{k=1}^n\|a_k\|\;\overset{\text{ by definition }}{=}\;\sum_{k=1}^\infty\|a_k\|$$

But we still need to show that, for any convergent series $\sum_{k=1}^\infty a_k$ in $X$, $$\bigg\|\sum_{k=1}^\infty a_k\bigg\|=\lim_{n\to\infty}\bigg\|\sum_{k=1}^na_k\bigg\|$$ in order to conclude that $$\bigg\|\sum_{k=1}^\infty a_k\bigg\|\leq \sum_{k=1}^\infty\|a_k\|.$$ (Of course, if $\sum_{k=1}^\infty a_k$ doesn't converge, then it makes no sense to talk about its norm in the first place, so we may as well assume it converges.)

The fact that $$\bigg\|\sum_{k=1}^\infty a_k\bigg\|=\lim_{n\to\infty}\bigg\|\sum_{k=1}^na_k\bigg\|$$ follows from the continuity of the norm function $\|\cdot\|:X\to\mathbb{R}$. Letting $$S_n=\sum_{k=1}^n a_k,$$ we have that $$\sum_{k=1}^\infty a_k=\lim_{n\to\infty} S_n.$$ For any topological spaces $A$ and $B$, continuous function $f:A\to B$, and convergent sequence $x_n\to x$ in $A$, we have that $f(x_n)\to f(x)$ in $B$. Thus, because the norm function $\|\cdot\|$ is continuous, we will get that $$\|\lim_{n\to\infty} S_n\|=\lim_{n\to\infty}\|S_n||$$ or in other words $$\bigg\|\sum_{k=1}^\infty a_k\bigg\|=\lim_{n\to\infty}\bigg\|\sum_{k=1}^na_k\bigg\|.$$

The reason the norm function is continuous is simply that the norm itself is what defines the topology on $X$. That is, the definition of $x_n\to x$ in $X$ is that $\|x_n-x\|\to 0$ in $\mathbb{R}$, and because the reverse triangle inequality tells us that $$\big|\|x\|-\|x_n\|\big|\leq\|x_n-x\|$$ we have that $$\lim_{n\to\infty}\big|\|x\|-\|x_n\|\big|=0$$ and hence $\lim_{n\to\infty}\|x_n\|=\|x\|$.

Note: I am assuming these sums in fact exist. If they fail to, the statement is meaningless.

You are not deducing anything about the $\infty$ case, which isn't even well-defined. The statement $$\left\|\sum\limits_{k=1}^\infty a_k\right\|\leq \sum\limits_{k=1}^\infty \|a_k\|$$ is defined as follows:

Let $x$ be the unique real number such that for all $\epsilon>0$ there exists some $N\in\mathbb N$ such that $$n\geq N\implies \left\|x-\sum\limits_{k=1}^n a_k\right\|<\epsilon$$ and let $y$ be the unique real number such that for all $\epsilon>0$ there exists some $N\in\mathbb N$ such that $$n\geq N\implies \left\|y-\sum\limits_{k=1}^n \|a_k\|\right\|<\epsilon.$$ Then $x\leq y$.

This makes no reference to $\infty$. Indeed, it works perfectly well to show that $$\left\|\sum\limits_{k=1}^n a_k\right\|\leq \sum\limits_{k=1}^n \|a_k\|$$ for all $n$ since this means that for any $\epsilon>0$ we have some $N$ such that if $n\geq N$ then $$x\leq y+\|x-y\|\leq y+\left\|\sum\limits_{k=1}^n a_k\right\|-\sum\limits_{k=1}^n \|a_k\|-\left\|x-\sum\limits_{k=1}^n a_k\right\|-\left\|y-\sum\limits_{k=1}^n \|a_k\|\right\|<y+2\epsilon$$ so we must have $x\leq y$.