We show that $\mathrm{ALG} \le (3/2) \mathrm{OPT} + 1$, and in some cases $\mathrm{ALG} \ge (3/2) \mathrm{OPT} -1/2$ even when the number of items is big enough. There is still a gap between the two results. Hope someone else can improve this result.
Consider an instance produced by your algorithm. Suppose the bins are opened in order
$b_1,\ldots, b_n$. Let $b_i$ be the bin such that $i=\max\{i\neq n\mid b_i\text{ contains exactly one item}\}$. Then we can conclude that for each bin $b_j$ where $j\le i$, there exists one item with size greater than 1/2. We call these items big items. In any optimal solution, we must also use $i$ bins to pack these big items. Suppose, besides these big items, there are also $k$ items $t_1,\ldots,t_k$ in the first $i$ bins. Observing that $t_1,\ldots,t_k$ are the smallest $k$ items and the $(k+1)$-th smallest item cannot be fit in any bin with a big item, we can conclude the items in the first $i$ bins are packed in an optimal way. In the rest of the proof, we will assume each bin, except the last one, contains at least two items.
Suppose the items used in step 1 are $h_1,\cdots, h_n$ in order and the rest items from small to big are $t_1,\ldots,t_m$. We have $t_1\le\cdots\le t_m\le h_n\le\cdots \le h_1$. Suppose bin $b_i$ has a space of $s_i$ that are not used. Recall that we assume each of $b_1,\ldots,b_{n-1}$ contains at least two items. We have $s_1+\cdots+s_{n-1}<t_1+\cdots+t_m+h_n$ due to step 2 of your algorithm, thus $s_1+\cdots+s_n<t_1+\cdots+t_m+1$. Note $s_1+\cdots+s_n+t_1+\cdots+t_m+h_1+\cdots+h_n=n$ and $s_i<h_{i+1}$ for $i<n$, we have $s_1+\cdots+s_n<(n+2)/3$. This means $\mathrm{ALG} \le (3/2) \mathrm{OPT} + 1$.
On the other hand, consider an instance with the following items (from small to large), where $\epsilon$ is a small number:
$$
\left\{\frac{1}{3}-6^k\epsilon,~~\frac{1}{3}-6^{k-1}\epsilon,~~\frac{1}{3}-6^{k-1}\epsilon, ~~ \text{ for } k = n, n-1,\ldots, 1\right\},
\\
\frac{1}{3}-\epsilon,
\\
\left\{\frac{1}{3}+2\times 6^{k-1}\epsilon,~~ \frac{1}{3}+3\times 6^{k-1}\epsilon,~~\frac{1}{3}+3\times 6^{k-1}\epsilon ~~ \text{ for } k = 1,\ldots, n\right\},
\\
\frac{1}{3}+2\times 6^n\epsilon.
$$
The optimal solution uses $2n+1$ bins:
$$
\left\{\frac{1}{3}-6^{k}\epsilon,~~ \frac{1}{3}+3\times 6^{k-1}\epsilon,~~\frac{1}{3}+3\times 6^{k-1}\epsilon ~~ \text{ for } k = 1,\ldots, n\right\},
\\
\left\{\frac{1}{3}-6^{k-1}\epsilon,~~ \frac{1}{3}-6^{k-1}\epsilon,~~\frac{1}{3}+2\times 6^{k-1}\epsilon ~~ \text{ for } k = 1,\ldots, n\right\},
\\
\frac{1}{3}-\epsilon, \frac{1}{3}+2\times 6^n\epsilon.
$$
But your algorithm uses $3n+1$ bins (each bin contains two items). Hence in this case $\mathrm{ALG} = (3/2) \mathrm{OPT} -1/2$.
