$\newcommand\de\delta$The maps 
$$\mu\mapsto\sqrt{\mathrm{KL}(\mu\|\nu)}$$
and 
$$\nu\mapsto\sqrt{\mathrm{KL}(\mu\|\nu)}$$
are not convex in general. 

Indeed, let $\mu_p:=p\de_0+(1-p)\de_1$, where $p\in(0,1)$ and $\de_a$ is the Dirac measure supported on $\{a\}$. 

Then the second partial derivative with respect to $p$ of $\sqrt{\mathrm{KL}(\mu_p,\mu_r)}$ at $(p,r)=(1/10,1/11)$ is $-7.17\ldots<0$. So, $\sqrt{\mathrm{KL}(\mu,\mu_r)}$ is not convex in $\mu$. 

Also, the second partial derivative with respect to $r$ of $\sqrt{\mathrm{KL}(\mu_p,\mu_r)}$ at $(p,r)=(1/10,1/9)$ is $-11.50\ldots<0$. So, $\sqrt{\mathrm{KL}(\mu_p,\nu)}$ is not convex in $\nu$. 

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You also asked: "If this is not true in general, does it exists a measure $\mu\in P(X)$ such that $\nu\mapsto \sqrt{\mathrm{KL}(\mu\|\nu)}$ is a convex map $P(X)\to \mathbb [0,+\infty]$? Or a measure $\nu$ such that $\mu\mapsto \sqrt{\mathrm{KL}(\mu\|\nu)}$ is convex?"

The answer to each of these two questions is yes, at least when $X=\{0,1\}$, say.  

For $p\in(0,1)$, let  
\begin{equation}
	F(p):=\sqrt{\mathrm{KL}(\mu_p,\mu_{1/2})},   
\end{equation}
\begin{equation}
	f(p):=F''(p)4 \mathrm{KL}(\mu_p,\mu_{1/2})^{3/2}, 
\end{equation}
\begin{equation}
	f_1(p):=f'(p)(1-p)^2 p^2.
\end{equation}
Then $f_1(1/2)=f'_1(1/2)=f''_1(1/2)=0$ and 
\begin{equation}
	f'''_1(p)=\frac{2+4 p(1-p)}{(1-p)^2 p^2}>0.
\end{equation}
It follows that $F''(p)\ge0$, so that $\sqrt{\mathrm{KL}(\mu,\mu_{1/2})}$ is convex in $\mu$. 


For $r\in(0,1)$, let  
\begin{equation}
G(r):=\sqrt{\mathrm{KL}(\mu_{1/2},\mu_r)},   
\end{equation}
and 
\begin{equation}
	g(r):=G''(r)4\mathrm{KL}(\mu_{1/2},\mu_r)^{3/2}.  
\end{equation}
Then $g(1/2)=g'(1/2)=g''(1/2)=g'''(1/2)=0$ and 
\begin{equation}
	g''''(1/2+h)\frac{(1 - 4 h^2)^4}{16}=9- 16 h^4 + 156 h^2  + 64 h^6
	>9- 1 + 156 h^2  + 64 h^6>0 
\end{equation}
if $|h|<1/2$.
It follows that $G''(r)\ge0$, so that $\sqrt{\mathrm{KL}(\mu_{1/2},\nu)}$ is convex in $\nu$.