Why is this 'Proof' by induction not valid?

With induction, you can only prove $S(n)$ is true for all positive integers $n$. However, even though $S(n)$ is true for arbitrarily large $n$, the statement "$S(\infty)$" does not follow from induction because $\infty$ is not a positive integer.

The same proof shows that the set of all positive integers is finite:

\begin{align} & \{1\} \text{ is finite.} \\ & \{1,2\} \text{ is finite.} \\ & \{1,2,3\} \text{ is finite.} \\ & \{1,2,3,4\} \text{ is finite.} \\ & \qquad \vdots \\ & \text{and so on.} \\ \text{Therefore } & \{1,2,3,4,\ldots\} \text{ is finite.} \end{align}

By induction you have proved that for all $n\in\mathbb Z^+$, $\displaystyle\sum_{k=1}^n\frac 1 k$ is finite, which is true. This is not the same as proving that $\displaystyle\sum_{k=1}^\infty\frac 1 k$ is finite...

I would add to the other comments that when you take the limit $<$ changes into $\le$. So by taking the limit you would get $\sum_{k=1}^\infty\frac{1}{k}\le\infty$, which is not particularly useful.

Other answers make the valid point that you can only deduce $( \forall n \in \mathbb N :P(n) )$ by induction, but not $ P(\infty) $ (though see footnote¹). There is, however, another problem in your case:

Your “$ P $” does not have the same meaning in “$ P(n)$” (where $ n\in \mathbb N $) as it does in “$ P(\infty) $”.

This is confusing, as the notation is the same, but an infinite sum is defined as a limit while a finite sum is defined inductively. Because of this, induction tells us nothing about $ P(\infty) $.

¹_{You can sometimes deduce $P(\omega)$ when using transfinite induction, but that is a different technique and a different story.}