Is there always an equivalent metric which is not complete?

As you noticed, all compatible metrics for compact metrizable spaces are complete.

For non-compact metrizable spaces, there is always a compatible metric which is not complete. For this, I will follow R. Engelking's General Topology (Exercise 4.3.E(d) and hint).

Let $X$ be a non-compact metrizable space with compatible metric $d$ bounded by $1$. Note that a metrizable space is compact iff it is countably compact, and so there is a decreasing sequence $\langle F_n \rangle_{n \in \mathbb{N}}$ of nonempty closed sets with empty intersection. Consider the mapping $\rho : X \times X \to \mathbb{R}$ defined by $$\rho ( x , y ) = \sum_n 2^{-n} \color{blue}{\left( | d ( x, F_n ) - d(y,F_n) | + \left[ \min \{ d(x,F_n) , d(y,F_N) \} \right] d(x,y) \right)}. \tag{1}$$ It can be shown that $\rho$ is a compatible metric for $X$ (this is largely because the part of (1) in blue defines a pseudometric on $X$ which is continuous as a function $X \times X \to \mathbb{R}$ for each $n$), and that $\operatorname{diam}_\rho (F_n) \leq 2^{-n}$ for each $n$. So if $\langle x_n \rangle_n$ is a sequence such that $x_n \in F_n$ for each $n$, then it is Cauchy with respect to $\rho$, but does not converge (since a limit would have to belong to $\bigcap_n F_n = \varnothing$).

Here is a stronger theorem.

Theorem: Let $X$ be a metrizable space and let $(x_n)$ be a sequence in $X$ with no accumulation point. Then $X$ embeds in a metrizable space $Y$ in which $(x_n)$ converges.

To answer your question from this theorem, note that if $X$ is not compact, it has a sequence $(x_n)$ with no accumulation point. If you then take $Y$ as in the theorem and restrict its metric to $X$, you get a metric on $X$ for which it is not complete, since $(x_n)$ is Cauchy but does not converge.

Proof of Theorem: Let $d_0$ be a metric inducing the topology of $X$ and let $Y$ be $X$ with one new point $*$ added. Let $d_1$ be a metric on the set $Z=\{x_n\}\cup\{*\}\subseteq Y$ such that $(x_n)$ converges to $*$ with respect to $d_1$. We define a metric $d$ on $Y$ as follows: for $a,b\in Y$, $d(a,b)$ is the infimum of all sums $$d_{i_0}(y_0,y_1)+d_{i_1}(y_1,y_2)+\dots+d_{i_n}(y_n,y_{n+1})$$ over all finite sequences of points $(y_0,\dots,y_{n+1})$ and sequences $(i_0,\dots,i_n)$ such that $y_0=a$, $y_{n+1}=b$, each $i_j$ is either $0$ or $1$, and if $i_j=0$ then $y_j,y_{j+1}\in X$ and if $i_j=1$ then $y_j,y_{j+1}\in Z$. Intuitively, $d(a,b)$ is the "shortest distance" between $a$ and $b$ where you are allowed to jump between points using either the metric $d_0$ or the metric $d_1$. It is easy to verify that $d$ is a pseudometric. Also, $d\leq d_1$ when restricted to $Z$, so $(x_n)$ converges to $*$ with respect to $d$. To verify that $d$ is in fact a metric (i.e., the distance between distinct points is positive) and that the restriction of $d$ to $X$ is equivalent to $d_0$, it suffices to prove the following claim:

Claim: For each $x\in X$, there exists $\epsilon>0$ such that $\{y\in X:d_0(x,y)<\epsilon'\}=\{y\in Y:d(x,y)<\epsilon'\}$ for all $\epsilon'\leq \epsilon$.

To prove this claim, note that since $(x_n)$ does not accumulate at $x$ in the topology of $X$, there exists $\epsilon_0>0$ such $d_0(x_n,x)>\epsilon_0$ for all $n$ such that $x_n\neq x$. Also, if $x\in Z$, then $x$ is isolated in $Z$ with respect to $d_1$ (since $(x_n)$ converges to $*$), so there exists $\epsilon_1>0$ such that $d_1(x,z)>\epsilon_1$ for all $z\in Z\setminus\{x\}$. Let $\epsilon=\min(\epsilon_0,\epsilon_1)$ (or just $\epsilon_0$ if $x\not\in Z$).

To show this value of $\epsilon$ works, suppose that $y\in Y$ and $d(x,y)<\epsilon'\leq\epsilon$; we must show that $y\in X$ and $d_0(x,y)<\epsilon'$ (note that the converse is trivial since $d\leq d_0$). So, let $(y_0,\dots,y_{n+1})$ and $(i_0,\dots,i_n)$ be such that $y_0=x$, $y_{n+1}=y$, and $$d_{i_0}(y_0,y_1)+d_{i_1}(y_1,y_2)+\dots+d_{i_n}(y_n,y_{n+1})<\epsilon'.$$ We may assume that the $y_j$ are all distinct and that the $i_j$ alternate between $0$ and $1$ (we can collapse consecutive terms with the same $i_j$ value using the triangle inequality). By our choice of $\epsilon_1$, there does not exist any $y_1\in Z$ such that $0<d_1(y_0,y_1)<\epsilon'$, so $i_0$ must be $0$. If $n>0$, this means $i_1$ must be $1$, so $y_1\in Z$. However, this is impossible, since $0<d_0(y_0,y_1)<\epsilon'$ and by our choice of $\epsilon_0$ there is no such $y_1$ that is in $Z$. Thus we can only have $n=0$, so our sequence is just $(x,y)$, $y\in X$, and $d_0(x,y)<\epsilon'$.