Let $\Omega$ be a domain in $\mathbb{C}^{m}$, $m\geq 1$. For each $a\in\Omega$, $i\in\{1,\ldots,m\}$, define $\Omega_{a,i}=\{z\in\mathbb{C}:a+ze_{i}\in\Omega\}$, where $e_{i}$ be the $i$-th standard basis element of $\mathbb{C}^{m}$. $\Omega$ is said to be convex in each variable if all of these $\Omega_{a,i}$'s are convex in $\mathbb{C}$. Does this force $\Omega$ to be polynomially convex $?$
A domain $D \subseteq \mathbb{C}^{m}$ is said to be polynomially convex if, for every compact subset $K$ of $D$, the polynomially convex hull of $K$, namely $\hat{K}=\{z\in\mathbb{C}^{n}:|p(z)|\leq \sup_{z\in K}|p(z)|$ for all polynomials$~p\}$ is contained in $D$.
The answer is negative, a counter-example is the union of two bi-disks: $$ \Omega= B_1\times B_1 \cup B_{2} \times B_{1/2}. $$ Here $B_r=\{z\in {\mathbb C}: |z|<r\}$.
Indeed, each coordinate slice of $\Omega$ is a round disk, hence, is convex. On the other hand, the polynomial hull of the closure $\overline{\Omega}$ contains the subset $$ H:=\{(t, t^{-1}): t\in {\mathbb C}, 1\le |t| \le 2\}. $$ (Apply the maximum principle to the holomorphic functions of one complex variable $p(t, t^{-1})$, where $p\in {\mathbb C}[z_1,z_2]$.) Clearly, $H$ is not contained in $\overline{\Omega}$. From this, you can easily see that $\Omega$ is not polynomially convex: For $r\in (0,1)$ consider compact subsets $$ K_r:= r \overline{\Omega}\subset \Omega. $$
For each $r$, $\hat{K}_r$ contains $rH$. Since for $r$ sufficiently close to $1$, $rH$ is not contained in $\Omega$, it follows that $\Omega$ is not polynomially convex.