Introduce ${\mathbf Z}_p$ as "formal" infinite base $p$ expansions where you add and multiply by carrying (any other description will probably take too long and not be concrete). Show them the series for $-1$ in ${\mathbf Z}_3$ is $2 + 2\cdot 3 + 2 \cdot 3^2 + 2 \cdot 3^3 + \cdots$ by adding 1 to that, carrying, and killing off a new term at each step so the sum is 0. Then emphasize the idea that in ${\mathbf Z}_p$ the number $p$ is small and redo the previous computation with geometric series: $2/(1 - 3) = 2/(-2) = -1$. Show ${\mathbf Z}_3$ contains a square root of 7: $1 + 3 + 3^2 + 2 \cdot 3^4 + 2 \cdot 3^5 + \cdots$.
(To explain why $p$ being prime is important, say the $p$-adic integers form an integral domain, and for a contrast you could define $Z_{10}$ in a similar way and say there is a number $x$ in ${Z}_{10}$ besides 0 and 1 satisfying $x^2 = x$: $x = 5 + 2\cdot 10 + 6\cdot 10^2 + 9\cdot 10^4 + 8\cdot 10^5 + \cdots$. Compute the first few digits of $x^2$ to check this works. This is related to the elementary school question of finding integers whose square ends in themselves: $5^2$ ends in 5, $25^2$ ends in 25, $625^2$ ends in 625, and so on. The 10-adic solution of $x^2 = x$ which I wrote the initial expansion of above packages all of this information into one number.)
You could introduce a topology on ${\mathbf Z}_p$ where numbers are close if a long string of initial digits are the same and make a metric from this too. Then indicate how this makes ${\mathbf Z}_p$ compact by a sequential argument, in the same spirit in which $[0,1]$ is compact by an argument with decimal expansions. A key new feature here, which those with experience only in real and complex analysis haven't seen before, is that ${\mathbf Z}_p$ is a compact ring. In ordinary geometry there are plenty of compact groups, but no compact rings.
I think a nice application of this compactness is the finiteness of integral solutions to certain equations. For example, $x^2 - 7y^2 = 1$ has an infinite number of integral solutions, but $x^3 - 7y^3 = 1$ has just two integral solutions: (1,0) and (2,1). One way to prove this finiteness is to use $3$-adic power series. By algebraic methods one can show that if integers $x$ and $y$ satisfy $x^3 - 7y^3 = 1$ then $x - y\sqrt[3]{7} = (2-\sqrt[3]{7})^n$ for some integer $n$; the solutions $(x,y) = (1,0)$ and $(2,1)$ correspond to $n = 0$ and $n = 1$, respectively. The point here is that when you expand $(2-\sqrt[3]{7})^n$ (say, by the binomial theorem when $n > 0$) you get a formula $a_n + b_n\sqrt[3]{7} + c_n\sqrt[3]{49}$ with integer coefficients $a_n, b_n, c_n$ and we want $c_n = 0$. That is a very strong constraint, since normally you don't expect $c_n = 0$. There is an exponential formula for $c_n$ in terms of $n$: $$ c_n = \frac{1}{21}\left(\sqrt[3]{7}(2 - \sqrt[3]{7})^n + \omega\sqrt[3]{7}(2 - \omega\sqrt[3]{7})^n + \omega^2\sqrt[3]{7}(2 - \omega^2\sqrt[3]{7})^n\right), $$ where $\omega$ is a cube root of unity. In the same way $a^x$ can be expanded as a real power series in $x$ when $a > 0$, the formula for $c_n$ above can be expanded into a $3$-adic power series in $n$ by interpreting $\omega$ and $\sqrt[3]{7}$ to be a cube root of unity and a cube root of 7 in the 3-adics. (Strictly speaking you need to pass to a finite extension of the 3-adics to pick up a cube root of unity, but let's gloss over that point.) Asking for $c_n$ to be 0 is then asking for $n$ to be an integral root of a 3-adic power series. Just as a nonzero analytic function on a closed (hence compact) disc in ${\mathbf C}$ has finitely many roots in the disc, a 3-adic power series that converges on ${\mathbf Z}_3$ has finitely many roots in ${\mathbf Z}_3$. Since ${\mathbf Z}$ is inside of ${\mathbf Z}_3$ this implies in particular that there are finitely many roots in ${\mathbf Z}$. To make this result effective (i.e., to know $n=0$ and $n=1$ are the only 3-adic roots of that power series), you need techniques to bound the number of $3$-adic roots of a $3$-adic power series, and that goes beyond the scope of your talk. :) Details are written, for instance, in section 6.4.7 of Henri Cohen's "Number Theory I: Tools and Diophantine Equations". This technique of proving effective finiteness theorems for integral solutions of Diophantine equations by $p$-adic methods goes back to work of Strassman and Skolem and later developments in this direction are due to Chabauty and Coleman.
Here is a cute application of $2$-adic continuity of polynomials (as a contrast to continuity of polynomials in the "usual" topology your audience will know). If we write out $\sqrt{1+x}$ as a power series it is $$ \sqrt{1+x} = \sum_{n \geq 0} \binom{1/2}{n}x^n = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \frac{5}{128}x^4 + \frac{7}{256}x^5 - \frac{21}{1024}x^6 + \cdots $$ and a striking feature is that the coefficients all have denominators that are powers of 2. It's not a surprise there is some power of 2 in the denominator considering a formula for the coefficient of $x^n$ is $$ \binom{1/2}{n} = \frac{(1/2)(1/2-1)(1/2-2)\cdots(1/2-n+1)}{n!}, $$ but why are there are no other primes in the denominator when you simplify the right side? One answer, which doesn't use anything $p$-adic, is to grind out the algebra and check that $$ \binom{1/2}{n} = \frac{(-1)^{n-1}}{2^{2n-1}}\left(\binom{2n-2}{n-1} - \binom{2n-2}{n}\right). $$ The binomial coefficients on the right are integers so the denominator is a power of 2. (That difference of binomial coefficients has a combinatorial interpretation as the $(n-1)$-th Catalan number.) Here is a slicker explanation of why the denominator is a power of 2: instead of directly seeing the power of 2 in the denominator of the rational number $\binom{1/2}{n}$, we will instead show for any prime $p$ other than 2 that $\binom{1/2}{n}$ is a $p$-adic integer, so it has no $p$ in its denominator as a reduced fraction. For odd $p$ we have the $p$-adic limit formula $\frac{1}{2} = \lim_{k \rightarrow \infty} \frac{p^k+1}{2}$, and the terms in that limit sequence are integers. The polynomial function $\binom{X}{n} \in {\mathbf Q}[X]$ is continuous in the $p$-adic topology, just as polynomials in ${\mathbf Q}[X]$ are continuous in the "usual" topology (in the reals), so by $p$-adic continuity $$ \binom{1/2}{n} = \lim_{k \rightarrow \infty} \binom{(p^k+1)/2}{n}. $$ Every binomial coefficient on the right is a positive integer for large $k$, so $\binom{1/2}{n}$ is a $p$-adic limit of integers and therefore is a $p$-adic integer. This means $\binom{1/2}{n}$ has no $p$ in its denominator. This holds for all odd $p$, so the only prime in the denominator of $\binom{1/2}{n}$ is 2.
The same basic idea shows for any nonzero rational number $r$ that the primes in the denominator of $\binom{r}{n}$ are limited to the primes in the denominator of $r$. For instance, the only primes in the denominator of $\binom{14/75}{n}$ are 3 and 5 because, with experience, one can see with $p$-adic limits that this fraction is $p$-adically integral for any $p$ other than 3 or 5. I doubt you're going to find a proof of that fact by some "explicit formula" method like the first proof I indicated for $\binom{1/2}{n}$ only having 2's in its denominator.