Is there a sufficiently reachable plausibility argument that $\pi$ is irrational?

But it must start to repeat after some digit, no matter how large, she retorted.

This sounds like a misunderstanding that you could do something about by showing her a concretely defined different irrational where it is clear to see that the digits cannot repeat, such as $$ \sum_{n=1}^\infty 10^{-n^2} $$ If you manage to convince her that is is possible for a number to have a decimal expansion that will never start (and continue!) repeating, it might become easier to accept that $\pi$ could be one of those numbers too.

I would try to convince her that most numbers are irrational and not even address $\pi$ - showing her that the question of "Do the digits of $\pi$ repeat?" is nontrivial (and, indeed, deep) is probably more valuable than convincing her that they don't, especially since convincing her of $\pi$'s irrationality could make it seem like $\pi$ is special when this is very much not a special property of $\pi$.

You can argue this fairly simply; the main thing to note is that an irrational plus a rational is irrational - this is easy to show via contradiction using integer ratios and is not too bad to show if you take "rational" as meaning "the digits repeat" either, so you can use whichever is likely to be more comfortable.

Right off the bat, once you get one irrational number, you have, in some sense, that at least $\frac{1}2$ the numbers are irrational, because if $c$ is any irrational number, then $x+c$ for $x$ rational is always irrational. A little more work can convince someone that $x+\alpha c$ for non-zero integer $\alpha$ will always be irrational and that the numbers of the form $x+\alpha c$ are all distinct based on the value of $(x,\alpha,c)$ - which seems to suggest that almost every number is irrational, since given any rational number, I can produce infinitely many irrational numbers!

Now, there's some elision in this argument - namely that it measures the size of sets in a vague way. The usual way to complete this argument would be to consider it mod $1$ (i.e. look at only fractional parts), then consider picking a random number in $[0,1)$ and think of the probability of hitting a rational - then, the formal thing that happens is that you can find infinitely many pairwise disjoint sets that are equally likely to hit as a rational, so the probability of hitting a rational must be zero.

As pointed out by @R.. in comments, one could rigorously define choosing a random real number in $[0,1]$ by rolling a fair ten sided die repeatedly to get its digits - this both can give an actual model to the argument I suggest and also could be used to give another proof that the probability of getting a rational is $0$ - it's not so hard to convince oneself that, for any fixed $n$ and $m$, the event that you roll $n$ digits and then a pattern with period $m$ occurs with probability $0$ - but then you'd need countable additivity to say that the probability of getting this for any $n$ and $m$ is also zero - and a student might rightfully balk at that reasoning since countable additivity is hard to motivate when uncountable additivity is clearly wrong.

A rigorous proof may not be possible at this level of mathematics. However, you may be able to get away with a more informal approach.

First, convince her that pi equals some generalized continued fraction, such as: $$ \pi = 3 + \frac{1^2}{6 + \frac{3^2}{6 + \frac{5^2}{6 + \frac{7^2}{\ddots}}}} $$ (or pick another example which converges faster?)

You can do this by computing partial sums and showing how they converge. Now, take some partial sums and try to simplify each one in turn. Write all integers as their prime factorizations, to make it obvious that they are getting progressively more complex and not "nicely cancelling." Now, you can credibly argue that you're not going to reach a regular fraction by following this process, so pi must be irrational.

This is not a "real" proof, and you should stress to her that there are additional pieces which you've left out. In particular, we are omitting a proof that the continued fraction really equals pi, and a proof that the continued fraction is irrational. You may also want to mention the fact that infinite sums need a formal definition.

For added rigor, use an alternating series, so that you can bracket pi from both above and below. This is arguably a "better" proof because it progressively removes from consideration fractions with larger and larger denominators, which means that you can't have "something funny" happen in the limit (cf. $0.999\ldots = 1$ and related limits). Another example suggested by PM 2Ring in the comments is the Wallis product: $$ \prod_{n=1}^\infty \bigg (\frac{2n}{2n-1} \cdot \frac{2n}{2n+1} \bigg ) = \frac{\pi}{2} $$ This brackets $\frac{\pi}{2}$ from above and below because the terms are alternately greater and less than one. It's also "perfectly obvious" that each partial product will have a progressively larger and larger power of two in the numerator (while the denominator is always odd).