I need to show that if $X_n \rightarrow X$ and $X_n \rightarrow Y$, then $X\overset{\text{a.s.}}{=}Y$ for convergence in probability, convergence almost surely, as well as for convergence in mean and quadratic mean ($\mathcal L^1$ and $\mathcal L^2$ convergence).
Convergence in Probability:
For any $\epsilon>0$ and for any $n\in\mathbb N$ we have
$$\begin{align} \mathbb P(|X-Y|\geq\epsilon) &\leq\mathbb P(|X-X_n|+|X_n-Y|\geq\epsilon)\\\\ &\leq\mathbb P\left((|X-X_n|\geq\epsilon/2)\cup(|X_n-Y|\geq\epsilon/2)\right)\\\\ &\leq\mathbb P(|X-X_n|\geq\epsilon/2)+\mathbb P(|X_n-Y|\geq\epsilon/2) \end{align}$$
so that
$$\mathbb P(|X-Y|\geq\epsilon)\leq\lim_{n\rightarrow\infty}\mathbb P(|X-X_n|\geq\epsilon/2)+\mathbb P(|X_n-Y|\geq\epsilon/2)=0$$
Since $$\{|X-Y|>0\}=\underbrace{\bigcup_{n=1}^\infty \underbrace{\left\{|X-Y|>\frac{1}{n}\right\}}_{=\emptyset}}_{=\emptyset}=\emptyset$$
we have that $\mathbb P\{|X-Y|>0\}=0$ and so $\mathbb P(X\ne Y)=0$. Hence $\mathbb P(X= Y)=1$ which means that $X\overset{\text{a.s.}}{=}Y$.
Convergence Almost Surely:
Since almost sure convergence implies convergence in probability, the result follows immediately from the last part. However, I'd like to show this without making use of that result. Since $X_n$ converges almost surely to both $X$ and $Y$ then $\mathbb P(\lim_{n\rightarrow\infty}X_n=X)=1$ and $\mathbb P(\lim_{n\rightarrow\infty}X_n=Y)=1$. From here it seems obvious to me that $X\overset{\text{a.s.}}{=}Y$ but I'm not sure how to show this formally.
Convergence in Mean:
$$\begin{align} \mathbb E(|X-Y|) &\leq\mathbb E\left(|X-X_n|+|X_n-Y|\right)\\\\ &=\mathbb E\left(|X-X_n|)+\mathbb E(|X_n-Y|\right) \end{align}$$
so
$$\mathbb E(|X-Y|)\leq\lim_{n\rightarrow\infty}\mathbb E(|X-X_n|)+\mathbb E(|X_n-Y|)=0$$
so $X\overset{\text{a.s.}}{=}Y$
Convergence in Quadratic Mean:
I tried continuing with the same logic but it's not the case that
$$ \mathbb E(|X-Y|^2)\leq\mathbb E\left(|X-X_n|^2+|X_n-Y|^2\right)$$
so I'm not sure how to proceed.
Is my reasoning correct for the first and third? How can I proceed with the other two?