Consider the Functional formula of the Riemann Zeta function
$$\zeta(s)=2^{s}\pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s)$$ Let $s=\frac{1}{2}+it$ for $t\in\mathbb{R}$ $$\zeta\left(\frac{1}{2}+it\right)=2^{\frac{1}{2}+it}\pi^{-\frac{1}{2}+it}\sin\left(\frac{\pi}{2}\Big(\frac{1}{2}+it\Big)\right)\Gamma\Big(\frac{1}{2}-it\Big)\zeta\Big(\frac{1}{2}-it\Big)$$ Lets say $\zeta\left(\frac{1}{2}+it\right)=0$ for some $t$.
Then $2^{\frac{1}{2}+it}\pi^{-\frac{1}{2}+it}\sin\left(\frac{\pi}{2}\Big(\frac{1}{2}+it\Big)\right)\Gamma\Big(\frac{1}{2}-it\Big)\zeta\Big(\frac{1}{2}-it\Big)$ at least one of the products must be $0$. In order to prove that $\zeta\left(\frac{1}{2}+it\right)=0$ if and only if $\zeta\left(\frac{1}{2}-it\right)=0$. We will show that $2^{\frac{1}{2}+it}\pi^{-\frac{1}{2}+it}\sin\left(\frac{\pi}{2}\Big(\frac{1}{2}+it\Big)\right)\Gamma\Big(\frac{1}{2}-it\Big)$ is never zero.
$\mathbf{Lemma\, 1.1}$ $$2^{\frac{1}{2}+it}\pi^{-\frac{1}{2}+it}\sin\left(\frac{\pi}{2}\Big(\frac{1}{2}+it\Big)\right)\Gamma\Big(\frac{1}{2}-it\Big)\neq 0\,\,,\,\forall t\in \mathbb{R}$$ $\mathbf{Proof:}$ We will show that they are non-zero term by term.
Obviously, $2^{\frac{1}{2}+it}\pi^{-\frac{1}{2}+it}$ cannot be zero, that clears out. $$\sin\left(\frac{\pi}{2}\Big(\frac{1}{2}+it\Big)\right)=\sin\Big(\frac{\pi}{4}+it\frac{\pi}{2}\Big)=\sin\Big(\frac{\pi}{4}\Big)\cos\Big(it\frac{\pi}{2}\Big)+\cos\Big(\frac{\pi}{4}\Big)\sin\Big(it\frac{\pi}{2}\Big)$$ $$=\frac{1}{\sqrt{2}}\cosh\Big(\frac{\pi}{2}t\Big)+\frac{i}{\sqrt{2}}\sinh\Big(\frac{\pi}{2}t\Big)$$ But we know $\cosh(t)$ is non-zero for all real $t$. This shows that $\sin\Big(\frac{\pi}{4}+it\frac{\pi}{2}\Big)\neq 0$
Now we only have to prove that $\Gamma\Big(\frac{1}{2}-it\Big)$ is non-zero and our proof will be done.
Recall the Weierstrass definition of the Gamma function $$\Gamma(s)=\frac{e^{-\gamma s}}{s}\prod_{k=1}^{\infty}\frac{e^{s/k}}{1+\frac{s}{k}}$$ Note that $x=0$ if and only if $|x|=0$. Plugging in $s=\frac{1}{2}+it$ and take absolute value on both sides gives $$\Big|\Gamma\Big(\frac{1}{2}+it\Big)\Big|=\left|\frac{e^{\frac{-\gamma}{2}-it\gamma }}{\frac{1}{2}+it}\right|\left|\prod_{k=1}^{\infty}\frac{e^{\frac{1}{2k}+\frac{it}{k}}}{1+\frac{1}{2k}+\frac{it}{k}}\right|=\frac{e^{-\frac{\gamma}{2}}}{|\frac{1}{2}+it|}\prod_{k=1}^{\infty}\frac{e^{\frac{1}{2k}}}{|1+\frac{1}{2k}+\frac{it}{k}|}$$ $$=\frac{e^{-\frac{\gamma}{2}}}{\sqrt{\frac{1}{4}+t^2}}\prod_{k=1}^{\infty}\frac{e^{\frac{1}{2k}}}{\sqrt{(1+\frac{1}{2k})^2+(\frac{t}{k})^2}}$$ $$\Rightarrow \Big|\Gamma\Big(\frac{1}{2}+it\Big)\Big|^2=\frac{e^{-\gamma}}{t^2+\frac{1}{4}}\prod_{k=1}^{\infty}\frac{e^{\frac{1}{k}}}{(1+\frac{1}{2k})^2+(\frac{t}{k})^2}$$ Remember the goal is to show that the entire thing is non-zero. We can show that $\prod_{k=1}^{\infty}\frac{e^{\frac{1}{k}}}{(1+\frac{1}{2k})^2+(\frac{t}{k})^2}$ is bounded above some $a>0$.
consider: $$a_k=\frac{(1+\frac{1}{2k})^2+(\frac{t}{k})^2}{e^{\frac{1}{k}}}$$ We have: $$\Big|\Gamma\Big(\frac{1}{2}+it\Big)\Big|^2=\frac{e^{-\gamma}}{t^2+\frac{1}{4}}\prod_{k=1}^{\infty}\frac{1}{a_k}$$ Consider the product $$\prod_{k=1}^{m}a_k=\prod_{k=1}^{m}\frac{(1+\frac{1}{2k})^2+(\frac{t}{k})^2}{e^{\frac{1}{k}}}=e^{-H_m}\prod_{k=1}^{m}\Big(1+\frac{1}{k}+\frac{1}{4k^2}+\frac{t^2}{k^2}\Big)$$ Using the taylor expansion of $e^x$ $$1+x\leq e^x$$ $$\Rightarrow 1+\frac{1}{k}+\frac{1}{4k^2}+\frac{t^2}{k^2}\leq e^{\frac{1}{k}+\frac{1}{4k^2}+\frac{t^2}{k^2}}$$ $$\Rightarrow \prod_{k=1}^m \Big(1+\frac{1}{k}+\frac{1}{4k^2}+\frac{t^2}{k^2}\Big)\leq \prod_{k=1}^m e^{\frac{1}{k}+\frac{1}{4k^2}+\frac{t^2}{k^2}}=e^{H _m+(t^2+\frac{1}{4})\sum_{k=1}^m\frac{1}{k^2}}$$ $$\leq e^{H_m +(t^2+\frac{1}{4})\zeta(2)}$$ Therefore $$\prod_{k=1}^{m}a_k \leq e^{-H_m}\cdot e^{H_m +(t^2+\frac{1}{4})\zeta(2)}=e^{(t^2+\frac{1}{4})\zeta(2)}$$ $$\Rightarrow \prod_{k=1}^{m}\frac{1}{a_k}\geq e^{-(t^2+\frac{1}{4})\frac{\pi^2}{6}}\,\,,\, \forall m\in\mathbb{N}$$ At the end, we have $$\Big|\Gamma\Big(\frac{1}{2}+it\Big)\Big|^2=\frac{e^{-\gamma}}{t^2+\frac{1}{4}}\prod_{k=1}^{\infty}\frac{1}{a_k}\geq \frac{e^{-\gamma-(t^2+\frac{1}{4})\frac{\pi^2}{6}}}{t^2+\frac{1}{4}}$$ Which is certainly greater than zero. $$ \Big|\Gamma\Big(\frac{1}{2}+it\Big)\Big|^2 \neq 0 \Rightarrow \Gamma\Big(\frac{1}{2}+it\Big) \neq 0$$
Hence the lemma is proved
$$\zeta\left(\frac{1}{2}+it\right)=2^{\frac{1}{2}+it}\pi^{-\frac{1}{2}+it}\sin\left(\frac{\pi}{2}\Big(\frac{1}{2}+it\Big)\right)\Gamma\Big(\frac{1}{2}-it\Big)\zeta\Big(\frac{1}{2}-it\Big)$$ $$\zeta\left(\frac{1}{2}+it\right)=\mathrm{(Not zero)}\zeta\Big(\frac{1}{2}-it\Big)$$ Finally, this shows that if $\zeta\left(\frac{1}{2}+it\right)=0$, then $\zeta\left(\frac{1}{2}+it\right)=0$ as well for all $t\in\mathbb{R}$.
Graphically, this tells us that the zeros of the Riemann zeta function on the critical line is reflective
$2$ Questions:
$(1)$ Is my proof correct?
$(2)$ I somehow felt like this is an excessive solution. Is there an alternate proof?
Your proof is correct, but the statement you are trying to prove is valid for any nontrivial zeros of $\zeta(s)$.
Note that when $s\in\mathbb R$ we have $\zeta(s)\in\mathbb R$, so by the principle of reflection we see that $\zeta(\overline s)=\overline{\zeta(s)}$. This indicates that $\zeta(s)=0$ iff $\zeta(\overline s)=0$. That is, all nonreal zeros of $\zeta(s)$ are symmetric about the real axis.