$X + Y \overset{\mathcal{D}}{=} X \Longrightarrow \mathbf{P}[Y = 0] =1$ 
Let $X$ and $Y$ be independent, real random variables. Show that $X + Y \overset{\mathcal{D}}{=} X$ implies that $\mathbf{P}[Y = 0] =1$.
Note: $U \overset{\mathcal{D}}{=} V$ means that the random variables $U$ and $V$ have the same distribution.

We know that a finite measure is fully determined by its characteristic function and vice versa.
Two random variables $U, V$ having the same distribution is equivalent to their pushforward measures $\mathbf{P}_U := \mathbf{P}\circ U^{-1}, \mathbf{P}_V$ being identical.
$$\phi_{\mathbf{P}_U}(t) = \int e^{i t x} \, d\mathbf{P}_U = \int e^{i t U} \, d\mathbf{P} =: \phi_U(t)$$
So $X + Y \overset{\mathcal{D}}{=} X$ is equivalent to $\phi_{X+Y} = \phi_{X}$.
Since $X, Y$ are independent, it must hold that $\phi_{X+Y} = \phi_X \phi_Y$, therefore
\begin{align*}
  \phi_X\phi_Y \overset{!}{=} \phi_X  \; \Rightarrow \; \phi_Y = 1 \; \Rightarrow \; 
   \int e^{i t Y }\, d \mathbf{P} = 1\,.
\end{align*}
One solution would be $Y = 0$ almost surely. But since a random variable is fully determined (almost surely) by its characteristic function, there cannot be another solution, so we conclude that $Y$ must be zero a. s. $\square$
Is everything ok? Thank you!
NOTE: It is not assumed that $X$ and $Y$ have finite variance.
 A: No, everything is not ok. You potentially divided by zero at a crucial step in your proof! In fact, the characteristic function of $X$ can vanish at certain $t$ values. For example, this happens when $X$ takes the values $0$ and $\pi$ with equal probability.
However, a modified argument can be used.

Claim. Let $X$ and $Y$ be independent random variables such that $X+Y$ and $X$ have the same distribution. Then $Y=0$ almost surely.

Proof.
Let $f(t)=\mathbb Ee^{itX}$, which is a continuous function of $t\in\mathbb R$. Since $f(0)=1$, applying continuity at $0$ yields an $\epsilon>0$ such that $|f(t)-1|<\tfrac12$ for all $|t|<\epsilon$. In particular, $|t|<\epsilon$ implies that $\mathbb Ee^{iXt}\not=0$.
By the hypotheses, we have that
$$
\mathbb Ee^{itX} \mathbb Ee^{itY}=\mathbb Ee^{itX},\qquad \forall t\in\mathbb R.
$$
When $|t|<\epsilon$, it follows that $\mathbb Ee^{itY}=1$. Taking real parts of both sides yields
$$
\mathbb E\cos(tY)=1,\qquad |t|<\epsilon.
$$
But this means that the non-negative random variable $1-\cos(tY)$ has mean zero, so that $1-\cos(tY)=0$ almost surely. Thus, $$Y\in \frac{2\pi}{t} \mathbb Z\text{ a.s.},\qquad \forall t\in (0,\epsilon).$$
Choosing $t_1,t_2\in (0,\epsilon)$ whose ratio is irrational leads to
$$
Y\in \frac{2\pi}{t_1} \mathbb Z\cap \frac{2\pi}{t_2} \mathbb Z=\{0\}\text{ a.s.},
$$
as desired.
