[proofplan]
We compare the median $m_F$ to the mean $\mathbb E[F]$ by contradiction on each side. If the median lies too far above the mean, then the median property forces the upper tail to have probability at least $1/2$, while the sub-Gaussian estimate makes that same tail strictly smaller than $1/2$. The lower-side argument is identical with the event $\{F\leq m_F\}$. Once the mean-median distance is bounded, the final tail estimate follows from the triangle inequality.
[/proofplan]
[step:Define the comparison radius and abbreviate the mean]
Let $\mu\in\mathbb R$ denote the expectation of $F$, so $\mu=\mathbb E[F]$. Define the radius $r\in(0,\infty)$ by
\begin{align*}
r:=\sigma\sqrt{2\log 4}.
\end{align*}
Then
\begin{align*}
2\exp\left(-\frac{r^2}{2\sigma^2}\right)=2\exp(-\log 4)=\frac{1}{2}.
\end{align*}
[/step]
[step:Rule out the median lying more than $r$ above the mean]
Suppose, toward a contradiction, that $m_F>\mu+r$. Define $s\in(r,\infty)$ by
\begin{align*}
s:=m_F-\mu.
\end{align*}
Since $\{F\geq m_F\}\subseteq \{F-\mu\geq s\}\subseteq \{|F-\mu|\geq s\}$, monotonicity of probability and the median property give
\begin{align*}
\frac{1}{2}\leq \mathbb P(F\geq m_F)\leq \mathbb P(|F-\mu|\geq s).
\end{align*}
Applying the assumed sub-Gaussian tail estimate at the value $s\geq 0$ gives
\begin{align*}
\mathbb P(|F-\mu|\geq s)\leq 2\exp\left(-\frac{s^2}{2\sigma^2}\right).
\end{align*}
Because $s>r$ and $\sigma>0$,
\begin{align*}
2\exp\left(-\frac{s^2}{2\sigma^2}\right)<2\exp\left(-\frac{r^2}{2\sigma^2}\right)=\frac{1}{2}.
\end{align*}
Combining the last three displays yields $\frac{1}{2}<\frac{1}{2}$, a contradiction. Therefore $m_F\leq \mu+r$.
[guided]
We want to prove that the median cannot sit too far to the right of the mean. Assume the contrary: $m_F>\mu+r$, where $\mu=\mathbb E[F]$ and $r=\sigma\sqrt{2\log 4}$. The amount by which the median exceeds the mean is the positive number $s:=m_F-\mu$, and the assumption gives $s>r$.
Now use the defining property of a median. Since $m_F$ is a median,
\begin{align*}
\mathbb P(F\geq m_F)\geq \frac{1}{2}.
\end{align*}
On the other hand, whenever $F\geq m_F$, we have $F-\mu\geq m_F-\mu=s$. Hence
\begin{align*}
\{F\geq m_F\}\subseteq \{F-\mu\geq s\}\subseteq \{|F-\mu|\geq s\}.
\end{align*}
By monotonicity of probability,
\begin{align*}
\frac{1}{2}\leq \mathbb P(F\geq m_F)\leq \mathbb P(|F-\mu|\geq s).
\end{align*}
The tail hypothesis applies at $s$ because $s>0$. It gives
\begin{align*}
\mathbb P(|F-\mu|\geq s)\leq 2\exp\left(-\frac{s^2}{2\sigma^2}\right).
\end{align*}
The strict inequality $s>r$ is the point where the endpoint issue is handled. Since the function $u\mapsto 2\exp(-u^2/(2\sigma^2))$ is strictly decreasing on $(0,\infty)$ and $\sigma>0$,
\begin{align*}
2\exp\left(-\frac{s^2}{2\sigma^2}\right)<2\exp\left(-\frac{r^2}{2\sigma^2}\right).
\end{align*}
By the definition of $r$,
\begin{align*}
2\exp\left(-\frac{r^2}{2\sigma^2}\right)=2\exp(-\log 4)=\frac{1}{2}.
\end{align*}
Thus $\mathbb P(|F-\mu|\geq s)<\frac{1}{2}$, contradicting the earlier lower bound $\mathbb P(|F-\mu|\geq s)\geq\frac{1}{2}$. Therefore $m_F\leq \mu+r$.
[/guided]
[/step]
[step:Rule out the median lying more than $r$ below the mean]
Suppose, toward a contradiction, that $m_F<\mu-r$. Define $s\in(r,\infty)$ by
\begin{align*}
s:=\mu-m_F.
\end{align*}
Since $\{F\leq m_F\}\subseteq \{\mu-F\geq s\}\subseteq \{|F-\mu|\geq s\}$, monotonicity of probability and the median property give
\begin{align*}
\frac{1}{2}\leq \mathbb P(F\leq m_F)\leq \mathbb P(|F-\mu|\geq s).
\end{align*}
Applying the assumed sub-Gaussian tail estimate at $s$ gives
\begin{align*}
\mathbb P(|F-\mu|\geq s)\leq 2\exp\left(-\frac{s^2}{2\sigma^2}\right)<2\exp\left(-\frac{r^2}{2\sigma^2}\right)=\frac{1}{2}.
\end{align*}
This contradiction proves $m_F\geq \mu-r$.
[/step]
[step:Combine the one-sided bounds to obtain the mean-median comparison]
The two preceding steps give
\begin{align*}
\mu-r\leq m_F\leq \mu+r.
\end{align*}
Equivalently,
\begin{align*}
|\mathbb E[F]-m_F|=|\mu-m_F|\leq r=\sigma\sqrt{2\log 4}.
\end{align*}
[/step]
[step:Transfer the sub-Gaussian tail from the mean to the median]
Fix $t\geq 0$. From the mean-median comparison, $|\mu-m_F|\leq r$. If $\omega\in\Omega$ satisfies $|F(\omega)-m_F|\geq t+r$, then the triangle inequality gives
\begin{align*}
|F(\omega)-\mu|\geq |F(\omega)-m_F|-|\mu-m_F|\geq t+r-r=t.
\end{align*}
Therefore
\begin{align*}
\{|F-m_F|\geq t+r\}\subseteq \{|F-\mu|\geq t\}.
\end{align*}
By monotonicity of probability and the assumed sub-Gaussian tail estimate at $t$,
\begin{align*}
\mathbb P(|F-m_F|\geq t+r)\leq \mathbb P(|F-\mu|\geq t)\leq 2\exp\left(-\frac{t^2}{2\sigma^2}\right).
\end{align*}
Substituting $r=\sigma\sqrt{2\log 4}$ gives
\begin{align*}
\mathbb P(|F-m_F|\geq t+\sigma\sqrt{2\log 4})\leq 2\exp\left(-\frac{t^2}{2\sigma^2}\right).
\end{align*}
This is the desired tail estimate around the median.
[/step]
