$\lim_{n \to \infty} x_n = 0$ and $\lim_{n \to \infty} f(x_n) = a$

For connectivity, apply the intermediate value theorem. Suppose that $x, y \in I$ and without loss of generality take $x < y$. Then there exist monotonically decreasing sequences $x_n$, $y_n$ converging to $0$ so $\lim\limits_{n \to \infty} f(x_n) = x$, $\lim\limits_{n \to \infty}f(y_n) = y$. Using terms from these sequences, one can form a decreasing sequence $z_n \to 0$ so that $\lim\limits_{k \to \infty}f(z_{2k}) = x$ and $\lim\limits_{k \to \infty}f(z_{2k+1}) = y$. Now fix $\alpha$ so $x < \alpha < y$. We can find a sequence of real numbers $\alpha_k$ so that for all $k$, $f(z_{2k}) \leq \alpha_k \leq f(z_{2k+1})$ and $\lim\limits_{k \to \infty}\alpha_k = \alpha$. Continuity of $f$ and the intermediate value theorem provide us a decreasing sequence $\tilde{z}_k \to 0$ so $f(\tilde{z}_k) = \alpha_k$. We conclude that $\alpha \in I$, so $I$ is connected.

To see that $I$ is closed, take $\alpha \in \mathbb{R}$ and a sequence $\alpha_k$ in $ I$ converging to $\alpha$. By the definition, we get decreasing sequences $x_{k, n}$ so that $\lim\limits_{n \to \infty} x_{k,n} = 0$ for all $k$ and $\lim\limits_{k \to \infty } f(x_{k,n}) = \alpha_k$ for all $k$. With this one can inductively construct a decreasing sequence $x_k$ converging to $0$ so that $|f(x_k) - \alpha_k| < \frac{1}{k}$. One can then see that $\lim\limits_{k \to \infty}f(x_k) = \alpha$, so $\alpha \in I$ completing the proof.

One should note that $I$ need not be a closed interval. The function $f(x) = \frac{1}{x} \sin(1/x)$ is an example where $I = \mathbb{R}$.