Hardy proved in 1914 that an infinity of zeros were on the critical line ("Sur les zéros de la fonction $\zeta(s)$ de Riemann" Comptes rendus hebdomadaires des séances de l'Académie des sciences. 1914).
Of course other zeros could exist elsewhere in the critical strip.
Let's exhibit the main idea starting with the Xi function defined by :
$$\Xi(t):=\xi\left(\frac 12+it\right)=-\frac 12\left(t^2+\frac 14\right)\,\pi^{-\frac 14-\frac{it}2}\,\Gamma\left(\frac 14+\frac{it}2\right)\,\zeta\left(\frac 12+it\right)$$
$\Xi(t)$ is an even integral function of $t$, real for real $t$ because of the functional equation (applied to $s=\frac 12+it$) :
$$\xi(s)=\frac 12s(s-1)\pi^{-\frac s2}\,\Gamma\left(\frac s2\right)\,\zeta(s)=\frac 12s(s-1)\pi^{\frac {s-1}2}\,\Gamma\left(\frac {1-s}2\right)\,\zeta(1-s)=\xi(1-s)$$
We observe that a zero of $\zeta$ on the critical line will give a real zero of $\,\Xi(t)$.
Now it can be proved (using Ramanujan's $(2.16.2)$ reproduced at the end) that :
$$\int_0^\infty\frac{\Xi(t)}{t^2+\frac 14}\cos(x t)\,dt=\frac{\pi}2\left(e^{\frac x2}-2e^{-\frac x2}\psi\left(e^{-2x}\right)\right)$$
where $\,\displaystyle \psi(s):=\sum_{n=1}^\infty e^{-n^2\pi s}\ $ is the theta function used by Riemann
Setting $\ x:=-i\alpha\ $ and after $2n$ derivations relatively to $\alpha$ we get (see Titchmarsh's first proof $10.2$, alternative proofs follow in the book...) :
$$\lim_{\alpha\to\frac{\pi}4}\,\int_0^\infty\frac{\Xi(t)}{t^2+\frac 14}t^{2n}\cosh(\alpha t)\,dt=\frac{(-1)^n\,\pi\,\cos\bigl(\frac{\pi}8\bigr)}{4^n}$$
Let's suppose that $\Xi(t)$ doesn't change sign for $\,t\ge T\,$ then the integral will be uniformly convergent with respect to $\alpha$ for $0\le\alpha\le\frac{\pi}4$ so that, for every $n$, we will have (at the limit) :
$$\int_0^\infty\frac{\Xi(t)}{t^2+\frac 14}t^{2n}\cosh\left(\frac {\pi t}4\right)\,dt=\frac{(-1)^n\,\pi\,\cos\bigl(\frac{\pi}8\bigr)}{4^n}$$
But this is not possible since, from our hypothesis, the left-hand side has the same sign for sufficiently large values of $n$ (c.f. Titchmarsh) while the right part has alternating signs.
This proves that $\Xi(t)$ must change sign infinitely often and that $\zeta\left(\frac 12+it\right)$ has an infinity of real solutions $t$.
Probably not as simple as you hoped but a stronger result!
$$-$$
From Titchmarsh's book "The Theory of the Riemann Zeta-function" p. $35-36\;$ and $\;255-258$ :
Best Answer
Note that if $1/2<\sigma<1, t \ne 0, \rho=\sigma+it$ we have $|\rho|^2=\sigma^2+t^2>(1-\sigma)^2+t^2=|1-\rho|^2$ so $\frac{2\sigma -1}{|\rho|^2} < \frac{2\sigma -1}{|1-\rho|^2}$ (since $2\sigma-1>0$) and by rearranging a little we get:
$\frac{2\sigma}{|\rho|^2} +\frac{2(1-\sigma)}{|1-\rho|^2}< \frac{1}{|\rho|^2}+\frac{1}{|1-\rho|^2}$
But now the expression above is symmetric in $\sigma, 1-\sigma$ so by interchanging them the inequality holds for $\sigma < 1/2$ too
Assume $\sum_\rho\frac{1}{\rho(1-\rho)}= \sum_{\rho}\frac{1}{|\rho|^2}$ and note that both sides are absolutely convergent hence we can regroup at will.
In particular, we rearrange the terms so we group $\rho, 1-\rho, \bar \rho, 1-\bar \rho$ together. If $\Re \rho =1/2$ this grouping has only two terms and appears twice (once for $\rho$ and once for $1-\rho$) while the sum is $\frac{1}{|\rho|^2}=\frac{1}{|1-\rho|^2}$ so the corresponding terms on RHS are equal.
If RH is false there is $\rho=\sigma+it, \sigma \ne 1/2, t > 0$ and the grouping above has $4$ terms each appearing twice (this time for $\rho, 1-\bar \rho$) with sum $\frac{1}{\rho(1-\rho)}+\frac{1}{\bar \rho(1-\bar \rho)}= \frac{2\sigma}{|\rho|^2} +\frac{2(1-\sigma)}{|1-\rho|^2}$ However in this case the corresponding terms on RHS are $\frac{1}{|\rho|^2}+\frac{1}{|1-\rho|^2}$ and by the result in the beginning we get a strict inequality.
Since by the grouping above we showed that LHS is a sum of positive terms, some equal with the corresponding ones on RHs and some strictly less we get the contradiction $\sum_\rho\frac{1}{\rho(1-\rho)} < \sum_{\rho}\frac{1}{|\rho|^2}$ so our supposition that RH is false cannot hold. Done!