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.

It's probably easier to picture it in $\mathbb{R}$: take $y := 0$, $\gamma_2 := 2$, $x_2 := 1$, $\gamma_1 := 1.5$ and $x_1 := 2.5$.
Can you see the issue there?
EDIT: There is no assumption you can make on $\gamma_1, \gamma_2$ to make the claim true.
Let $y \in \mathbb{R}$ and $\gamma_2 > 0$ be fixed. Pick $x_2 := \gamma_2$. Then for all $\gamma_1 > 0$, there exists $x_1 \in \overline{B}(x_2,\gamma_1) \setminus \overline{B}(y,\gamma_2)$, just by taking $x_1 := x_2 + \gamma_1 = \gamma_2 + \gamma_1$.
