I am not sure that this is a research level question.
Remark 10.9.4 in the book "A course in metric geometry" by Burago, Burago, Ivanov claims the following. 


Let $X$ be a finite dimensional Alexandrov space with curvature bounded below. Fix $p\in X$ and $\varepsilon> 0$. Then there exists $r>0$ such that for any two points $x,y$ such that $|px|,|py|<r$ one has
$$|\tilde\measuredangle xpy -\measuredangle xpy|<\varepsilon.$$


**I would be happy to have a detailed proof of this result or a reference to it.** 


Thank you.



Angle estimate in Alexandrov spaces
Partly following the advise in Anton Petrunin's answer, let me present a proof in greater detail.


We prove the statement in the case of non-negative curvature. Assume the contrary. Then there exist $\delta >0$ and sequences $x\_n,y\_n\to p$ such that $0<|px\_n|\leq |py\_n|\to 0$, and 
\begin{eqnarray}\label{0}
\tilde\measuredangle x\_npy\_n< \measuredangle x\_npy\_n-\delta.
\end{eqnarray}
We denote for brevity 
$$r\_n:=|px\_n|,\, R\_n:=|py\_n|,\, \alpha\_n:=\measuredangle x\_npy\_n, \tilde\alpha:=\tilde\measuredangle x\_npy\_n.$$
Thus
$$r\_n\leq R\_n\to 0,\, \tilde\alpha\_n<\alpha\_n-\delta.$$


Choosing a subsequence, we may assume that there exists a geodesic $\gamma$ starting at $p$ such that the angle between its direction at $p$ and the direction of $px\_n$ is at most a small $\omega$ to be chosen later.
Let $z\_n\in \gamma$ be such that $|pz\_n|=r\_n$.


We have
\begin{eqnarray}\label{1}
|x\_nz\_n|=\sqrt{2r\_n^2(1-\cos\tilde\measuredangle x\_npz\_n)}=2r\_n \sin{(\frac{\tilde\measuredangle x\_npz\_n}{2})}\leq r\_n\cdot \omega.
\end{eqnarray}
Let us show that the angles $\tilde \measuredangle z\_npy\_n$ and $\tilde \measuredangle x\_npy\_n$ are very close to each other. We have
\begin{eqnarray\*}
|x\_ny\_n|^2=r\_n^2+R\_n^2-2r\_nR\_n\cos\tilde\measuredangle x\_npy\_n,\\
|z\_ny\_n|^2=r\_n^2+R\_n^2-2r\_nR\_n\cos\tilde\measuredangle z\_npy\_n.
\end{eqnarray\*}
Subtracting the two equalities we get 
\begin{eqnarray\*}
2r\_nR\_n(\cos\tilde\measuredangle z\_npy\_n-\cos\tilde\measuredangle x\_npy\_n)=(|z\_ny\_n|-|x\_ny\_n|)(|z\_ny\_n|+|x\_ny\_n|).
\end{eqnarray\*}


This and (\ref{1}) imply
\begin{eqnarray\*}
|\cos\tilde\measuredangle z\_npy\_n-\cos\tilde\measuredangle x\_npy\_n|\leq \omega \frac{|z\_ny\_n|+|x\_ny\_n|}{2R\_n}\leq \omega \frac{r\_n+R\_n}{R\_n}\leq 2\omega.
\end{eqnarray\*}
Since $|\arccos a-\arccos b|\leq C|a-b|^{1/2}$ for some fixed $C\geq 1$ and any $a,b\in [0,\pi]$, we obtain
\begin{eqnarray}\label{2}
|\tilde\measuredangle z\_npy\_n-\tilde\alpha\_n|\leq C\sqrt{2\omega}.
\end{eqnarray}
On the other hand
\begin{eqnarray\*}
\measuredangle z\_npy\_n-\tilde\measuredangle z\_npy\_n\geq (\measuredangle x\_npy\_n-\measuredangle z\_npx\_n)-(\tilde\alpha\_n+C\sqrt{2\omega})\geq\\
\measuredangle x\_npy\_n-\tilde\alpha\_n-(\omega+C\sqrt{2\omega})\geq \delta -(\omega+C\sqrt{2\omega}).
\end{eqnarray\*}
Now let us choose $\omega$ so that $\omega+C\sqrt{2\omega}<\delta/2$. Then we get
\begin{eqnarray}\label{3}
\measuredangle z\_npy\_n-\tilde\measuredangle z\_npy\_n> \delta/2.
\end{eqnarray}


Thus we reduced the problem to the case when the points $z\_n$ lie on the same geodesic $\gamma$ and $|pz\_n\leq |py\_n|$. Now choose point $w\_n\in \gamma$ such that $|pw\_n|=|py\_n|$.
We have $\tilde\measuredangle w\_npy\_n\leq \tilde\measuredangle z\_npy\_n$. Hence by (\ref{3}) we have
\begin{eqnarray}\label{4}
\measuredangle w\_npy\_n-\tilde\measuredangle w\_npy\_n> \delta/2.
\end{eqnarray}
Thus now $|pw\_n|=|py\_n|$. Like in the previous step we can find another geodesic $\hat\gamma$ such that its direction at $p$ is very close to the directions at $p$ of the geodesics $py\_n$.
We choose points $t\_n\in \hat \gamma$ such that $|pt\_n|=|py\_n|$. Repeating the argument as above we get for $n\gg 1$
$$\measuredangle w\_npt\_n-\tilde\measuredangle w\_npt\_n> \delta/4.$$
But $\measuredangle w\_npt\_n$ equals the angle between $\gamma$ and $\hat\gamma$, and $\tilde\measuredangle w\_npt\_n$ converges to this angle by definition. This is a contradiction. QED


