Here is the sketch of a proof. All we need is that every continuous weak order on each $X_i$ admits a continuous utility representation. One sufficient condition is that each $X_i$ is a connected separable topological space by a theorem of Eilenberg. A proof of Eilenberg's theorem is given in Debreu's book Theory of Value. Debreu assumes the domain there to be Euclidean, but the proof easily generalizes to the present setting. One can also use the assumption that each $X_i$ is a second-countable topological space (for metrizable spaces implied by separability). Debreu showed that continuous utility representations exist in that setting too, but the proof is considerably harder.
Define $\succeq_i$ on $X_i$ by $x_i\succeq_i y_i$ if $x_iz\succeq y_iz$ for some $z$. Clearly, this defines a continuous weak order and the $z$ used does not matter by singleton separability. There exists a continuous utility function $u_i:X_i\to\mathbb{R}$.
Next, we can show that $x_i~\sim_iy_i$ for $i=1,\ldots,N$ implies $x\sim y$. By repeated applications of singleton separability we get $(x_1,y_2,y_3,\ldots,y_N)\succeq (y_1,y_2,y_3,\ldots,y_N)$, $(x_1,x_2,y_3,\ldots,y_N)\succeq (x_1,y_2,y_3,\ldots,y_N)$, and so on, until we have $(x_1,x_2,\ldots,x_{N-1},x_N)\succeq(x_1,x_2,\ldots,x_{N-1},y_N)$. By transitivity, $x\succeq y$. Similarly, $y\succeq x$, and, therefore $x\sim y$.
Let $U=\prod_{i=1}^N u_i(X_i)$, a product of (possibly unbounded) intervals. Define $\succeq^*$ on $U$ by $u\succeq^* u'$ if for some $x,y\in X$ one has $u=\big(u_1(x_1),\dots,u_N(x_N)\big)$, $u'=\big(u_1(y_1),\dots,u_N(y_N)\big)$, and $x\succeq y$. By what we have just shown, $\succeq^*$ contains, together with $u_1,\ldots, u_N$, the same information as $\succeq$. It is also obvious that $\succeq^*$ is strictly monotone on $U$. If a utility representation $W:U\to\mathbb{R}$ of $\succeq^*$ exists, it must be strictly increasing. We are done if we can show a continuous such representation exists. Using any of the mentioned utility representation theorems, it suffices to show that $\succeq^*$ is continuous.
So let $u\succ^* u'.$ There exist $x,y\in X$ such that $u=\big(u_1(x_1),\dots,u_N(x_N)\big)$, $u'=\big(u_1(y_1),\dots,u_N(y_N)\big)$, and $x\succ y$. Since $\succeq$ is continuous and by the definition of the product topology, there exist open neighborhoods $V_i\ni x_i$ and $W_i\ni y_i$ such that $x'\in\prod_{i=1}^N V_i$ and $y'\in\prod_{i=1}^N W_i$ implies that $x'\succ y'$. Let $V$ be the interior of $\prod_{i=1}^N u_i(V_i)$ and $W$ the interior of $\prod_{i=1}^N u_i(W_i)$. Then for $u''\in V$ and $u'''\in W$ one has $u''\succ^* u'''$, so $\succeq^*$ is, indeed, continuous.
