What are the unknown steps to show $\mathbf{F}=-\nabla\psi$? Given: $\mathbf{F}=F(r)\ \mathbf{\hat{r}}$ is a spherically symmetric radial vector field which is continuous everywhere except at origin.
To prove: $\mathbf{F}=-\nabla \psi$
Proof:
\begin{align}
\mathbf{F} \cdot \mathbf{\hat{r}}&=F(r)\ \mathbf{\hat{r}} \cdot \mathbf{\hat{r}}=F(r)=\dfrac{d(\int F\ dr)}{dr}\\ &=\dfrac{d\psi}{dr}=\nabla \psi \cdot \mathbf{\hat{r}}
\end{align}
From here we cannot straight away deduce that: $\mathbf{F}=\nabla\psi$
UNKNOWN STEPS ???
$$\mathbf{F}=-\nabla\psi$$
$\mathbf{F}$ is conservative.

Question: From what steps (unknown to me) can we get $\mathbf{F}=-\nabla\psi\ $?

 A: Let's write 
$$
\newcommand{\rr}{{\mathbf{\hat{r}}}}
\newcommand{\th}{{\mathbf{\hat{\theta}}}}
\newcommand{\FF}{{\mathbf{F}}}
\FF(r, \theta) = F(r) \rr + 0 \th
$$
And define 
$$
\psi(r, \theta) = -\int_0^r F(t) ~ dt
$$
Then we have
\begin{align}
\frac{d\psi}{dr}(r, \theta) &= -F(r)\\
\frac{d \psi}{d\theta}(r, \theta) &= 0\\
\nabla \psi (r, \theta) 
&= \frac{1}{r} \frac{d \psi}{d\theta} \th  + \frac{d \psi}{dr}\rr\\
&= \frac{1}{r} 0 ~ \th  + -F(r)\rr\\
&= -F(r)\rr \tag{*}\\
\end{align}
so that $\nabla \psi$ is again a radial vector field. Two radial vector fields are equal if the coefficient of $\rr$ is the same in each. 
As you observe,
\begin{align}
\FF(r, \theta) \cdot \rr 
&= F(r) \rr  \cdot \rr \\
&= F(r) \\
&= -\frac{d\psi}{dr} (r, \theta) \\
&= -\nabla \psi (r, \theta) \cdot \rr
\end{align}
Alternatively, you could just look at equation (*) and compare it to the definition of $\FF$ and see that $\FF = -\nabla \psi$. 
Summary: 


*

*You needed to include a minus-sign in the definition of $\psi$ to get the answer you were asked for.

*You needed to make the observation that the gradient of $\psi$ was a radial field, and then the computation you did would suffice. 
A: You have a differential 
equation 
here in $-\operatorname{grad}\psi=\mathbf F=f(\mathbf r) \frac{\bf r}{\|\mathbf r\|}$, where $f$ is a scalar function and $\mathbf F$ a vector function. 
One way to solve it for $\psi$ is by guessing a solution and then checking if its right. 
By definition :
$$\mathbf r\cdot\operatorname{grad}\psi=\lim_{t\to0}\frac{\psi(\mathbf x+t\mathbf r)-\psi(\mathbf x)}t$$ 
where $t$ is a scalar and $\mathbf x, \mathbf r$ vectors. 
From this follows :
$$\operatorname{grad}(\|\mathbf r\|)=\frac{\mathbf r}{\|\mathbf r\|}$$
And chain rule : 
$$\operatorname{grad}(f(g(\mathbf r)))=\frac{\mathrm df}{\mathrm dg}\operatorname{grad}(g)$$
Now guessing $\psi(\mathbf r)=-\left(\int f(\mathbf r)\,\mathrm d\|\mathbf r\|\right)\cdot\|\mathbf r\|$
we get $-\operatorname{grad}\psi=f(\mathbf r)\frac{\mathbf r}{\|\mathbf r\|}$ so this is the right solution to the differential equation. 
