STFT uniform continuity

Question

I want to show that short-time Fourier transform of $f \in L^2$ w.r.t $g \in L^2$ \begin{align*} \mathcal{V}_{g} f (x, \omega) = \int\limits_{\mathbb{R}}^{} f(t) \overline{g(t-x)} e^{-2 \pi i t \omega} dt = \langle f, M_{\omega} T_x \, g \rangle \end{align*} (where $M_{\omega}$ is modulation and $T_x$ translation) is uniformly continuous on $\mathbb{R} ^2$. My attempt: \begin{align*} \lvert \mathcal{V}_{g} f (x, \omega) - \mathcal{V}_{g} f (x', \omega ')\rvert & = \lvert \langle f, M_{\omega} T_x g - M_{\omega '} T_{x'} g \rangle \rvert \\ & \leq \lVert f \rVert _2 \, \lVert M_{\omega } T_{x} g - M_{\omega '} T_{x'} g \rVert _2 \end{align*} and I want to use \begin{align*} \lVert T_x h - h \rVert _2 \rightarrow 0 \; \; \; \text{as} \; \; \; \; x \rightarrow 0 \end{align*} \begin{align*} \lVert M_{\omega } h - h \rVert _2 \rightarrow 0 \; \; \; \text{as} \; \; \; \; \omega \rightarrow 0 \end{align*} so I tried estimating \begin{align*} \lVert M_{\omega } T_{x} g - M_{\omega '} T_{x'} g \rVert _2 & \leq \lVert M_{\omega } T_{x} g - M_{\omega '} T_{x} g \rVert _2 + \lVert M_{\omega '} T_{x} g - M_{\omega '} T_{x'} g \rVert _2 \\ & = \lVert M_{\omega } T_{x} g - M_{\omega '} T_{x} g \rVert _2 + \lVert T_{x-x'} g - g \rVert _2 \end{align*} and now I am struggling to estimate first term in the last line ($\lVert M_{\omega } T_{x} g - M_{\omega '} T_{x} g \rVert _2$) in such a way that I can apply the limits above. I would appreciate any help.