Let $U, V$ be two neighborhoods of $x_0$. Prove that $U\cap V$ is also a neighborhood of $x_0$ 
Let $U, V$ be two neighborhoods of $x_0$. Prove that $U\cap V$ is also a neighborhood of $x_0$

My attempt
I know that $I$ is a neighborhood of $y_0$ if $B(y_0,r)\subseteq I$ for some $r$ where $$B(y_0,r)=\left\{y\in\mathbb R:|y-y_0|<r\right\}$$


*

*If  $U\subset V$ or  $V\subset U$, I can prove it by considering the $B$ of the smallest neighborhood.

*If $U=V$ I can prove it because $B_V=B_U=B_{U\cap V}$


If neither 1. nor 2. apply, (the following came to my mind as I was writing this question) I think I can prove it by considering $B=B_V\cap B_U$. Am I right?
 A: Since $U$ and $V$ are neighbourhoods of $x_0$, there exist $r,r'>0$ such that $B(x_0, r) \subseteq U$ and $B(x_0, r') \subseteq V$. What you should do now is prove that $B(x_0, R) \subseteq U \cap V$ for some suitable choice of $R>0$.
Hover your cursor over this box for a bigger hint:

 I suggest taking $R = \mathrm{min} \{ r,r' \}$.

A: A purely topological proof
Definition 1. (One of the many equivalent definitions) A topology on  a set $X$ is a subset $\mathscr{T}$ of the power set $\mathscr{P}(X)$, having the following properties


*

*$X,\emptyset\in\mathscr{T}$

*For every family $\{A_i\}$ of subsets of $X$ in $\mathscr{T}$ 
$$
   \bigcup_i A_i\in\mathscr{T}
   $$

*For every finite family $\{A_i\}$ of subsets of $X$ in $\mathscr{T}$ 
$$
   \bigcap_i A_i\in\mathscr{T}
   $$


The members of $\mathscr{T}$ are the open sets of the so defined topology, and the ordered couple $(X,\mathscr{T})$ is a topological space. When it is clear of what kind of topology we are dealing with, its explicit mention can be dropped so we simply speak of a topological space $X$.
Definition 2. A neighborhood $V$ of a given set $P\subset X$ of a topological space $X$ is subset of $X$ which contains an open set $O$ such that $V\supset O\supset P$.
Lemma If $V$ and $U$ are neighborhoods of a point $x_0$ of a topological space $X$, then also $V\cap U$ is a neighborhood of $x_0$.
Proof. If $O_V\subset V$ and $O_U\subset U$ are open sets containing $x_0$, their intersection $O_V\cap O_U$ 


*

*is not empty since contains $x_0$ and

*is an open set by property 3 of definition 1 above and

*is a subset of $V\cap U$ by construction.


By definition 2, $V\cap U$ is a neighborhood of $x_0\in X$. $\blacksquare$
Few notes


*

*This neighborhood property follows very quickly from the definition of neighborhood itself and the definition of topology: there is no need of defining balls or other metrical "devices". 

*In $\mathbb{R}^n$, $n\geq 1$ (and not only in it) equipped by its "natural" topology, i.e. the one defined by its Euclidean distance function, you can consider only ball instead of the general open sets, since the family $\{B(x_0,R)\subset \mathbb{R}^n\,|\,x\in\mathbb{R}^n , R\in\mathbb{R}_+\}$ is a topological basis for $\mathbb{R}^n$. The proofs given by Clive Newstead and Kanu Kim implicitly rely on this fact. 

A: There exist such radii $r_U$ and $r_V$ for $U$ and $V$. Then just take the minimum of them: $r = \min(r_U, r_V)$ whence $B(x_0,r) \subseteq B(x_0, r_U)\cap B(x_0, r_V) \subseteq U \cap V$, proving $U\cap V$ is also a neighbourhood of $x_0.$
This method can be applied for finitely many open neighbourhoods, but not for infinitely many, since the infimum of infinitely many positive numbers could be zero.
