Showing that a measure is lower continuous. Here is the problem and my attempt.

Let $\mu$ be a measure defined on sets $E_1, E_2, \ldots$ such that $E_{n+1} \subset E_n$ for all $n$. Additionally, let $\mu(E_1) \lt \infty$. Show that $$\mu\left(\bigcap_{n = 1}^\infty E_n\right) = \lim_{n\to\infty} \mu (E_n).$$

We know that, for all $k$, $E_k \supset \bigcap_{n = 1}^\infty E_n$ so we can state that $\mu(E_k) \leq \mu\left(\bigcap_{n = 1}^\infty E_n\right)$ and then take limits to find $$\lim_{n\to\infty}\mu(E_n) \leq \mu\left(\bigcap_{n = 1}^\infty E_n\right)$$
Similarly, for all $k$, $\mu(\bigcap_{n = 1}^k E_k) \leq \lim_{n\to \infty} \mu(E_n)$. Taking limits gives
$$\lim_{k\to\infty}\mu\left(\bigcap_{n = 1}^k E_n\right) = \mu\left(\bigcap_{n = 1}^\infty E_n\right) \leq \lim_{n\to\infty}\mu(E_n).$$
I do not feel like my treatment of limits was appropriate. Is this proof correct?

Actually, the fact that $E_k\supseteq \bigcap_{n=1}^{+\infty}E_n$ tells you that $\mu(E_k) \geq \mu(\bigcap_{n=1}^{+\infty}E_n)$ and not the other way round. And then in your last passage you use the equality $$\lim_{k\to +\infty} \mu(\bigcap_{n=1}^k E_k) = \mu(\bigcap_{n=1}^{+\infty} E_n)$$ that is precisely what you want to prove, since $\bigcap_{n=1}^k E_n = E_k$.
However is not too hard to adjust these things, for example like this:  we need to show that 
$$ \lim_{k\to\infty} \mu(E_k) = \mu(\bigcap_{n=1}^{+\infty} E_n) $$
and this is equivalent to 
$$ \mu(E_1)-\lim_{k\to\infty} \mu(E_k) = \mu(E_1)-\mu(\bigcap_{n=1}^{+\infty} E_n) $$
that in turn is equivalent to
$$  \lim_{k\to +\infty}\mu(E_1\setminus E_k) = \mu(\bigcup_{n=1}^{+\infty}E_1\setminus E_n)$$
now write $A_n=E_n\setminus E_{n+1}$: then the $A_i$ are mutually disjoint and $E_1\setminus E_n = \bigcup_{i=1}^{n-1}A_i$ so that 
$$ \mu(\bigcup_{n=1}^{+\infty}E_1\setminus E_n) = \mu(\bigcup_{n=1}^{+\infty}A_n) = \sum_{n=1}^{+\infty} \mu(A_n)=\lim_{k\to +\infty}\sum_{i=1}^{k-1}\mu(A_i)=\lim_{k\to+\infty}\mu(E_1\setminus E_k) $$
