Does $x_1$ belong to $\overline{B}(y, \gamma_2)$? Let $(X, d)$ be a metric space and let $0<\gamma_1<\gamma_2$. Let $x_1, x_2\in X$ and $y\in X$ be fixed. Assume that
$$ x_2\in \overline{B}(y, \gamma_2)\quad\text{ and }\quad d(x_1, x_2)\le \gamma_1.$$
It is true that $x_1\in\overline{B}(y, \gamma_2)$?
My attempt: About me the answer is yes, since as $x_2\in \overline{B}(y, \gamma_2)$ and its distance from $x_1$ is less than $\gamma_2$ is trivial that also $x_1\in\overline{B}(y, \gamma_2)$.
On the contrary, during my calculus class another guy said that the answer is no. This is is tentative proof:
$$d(x_1, y)\le d(x_1, x_2)+d(x_2, y)\le \gamma_1 +\gamma_2\le 2\gamma_2. $$
$\bf{EDIT:}$ If not, which further assumptions are needed on $\gamma_1, \gamma_2$ to make $x_1\in\overline{B}(y, \gamma_2)$ true?
Could someone please help me to understand who is doing it right?
Thank you.

The second method gives the correct answer.
You have $|x_2-y|\le \gamma_2$ and $|x_2-x_1|\le \gamma_1 <\gamma_2$, therefore $|x_1-y|\le \gamma_2$, but this isn't true.
If $d(x_1,y)+d(x_2,y)\le\gamma_2$, then $x_1\in\overline{B}(y, \gamma_2)$, otherwise some $x$ are inside the ball, and some $x$ are not.
