Here is an extract of my Symplectic Geometry course notes.
Remark 2.57. Let $M$ be a smooth manifold, $Q\subseteq M$ a compact submanifold and $\omega_0,\omega_1\in\Omega^2(M)$ symplectic forms such that for all $t\in[0,1]$ and $q\in Q$, $\omega_t:=(1-t)\omega_0+t\omega_1$ is non degenerate in $q$ (hence on $T_qM$). Then there exist an open neighbourhood $U$ of $Q$ such that for all $t\in[0,1]$, $\omega_t$ is symplectic on $U$.
If this were false infact, there would be a sequence $(t_j)_{j\in\mathbb{N}}$ in $[0,1]$ and one $(m_j)_{j\in\mathbb{N}}$ in $U_{1/j}$ such that $\omega_{t_j}$ is degenerate in $m_j$, but $[0,1]$ and $Q$ are compacts, so we can extract a subsequence $j_k$ with $k\in\mathbb{N}$ such that the two sequences converge.$$t_{j_k}\xrightarrow[k\to\infty]{}t_\infty\qquad\qquad m_{j_j}\xrightarrow[k\to\infty]{}m_\infty$$
Now, since $m_{j,k}\in U_{1/j_k}$, $m_\infty\in U_{j_k}$ for all $k$ (otherwise we can find an open neighborhood of $m_infty$ disjoint with all the $m_{j_k}$ for all $k$ large enough). Thus $m_\infty\in\bigcap_kU_{1/j_k}=Q$. Now in every local chart centred in $m_\infty$, the matrix representing $\omega_{t_\infty}$ in $m_\infty$ is the limit of the matrices representing $\omega_{j_k}$ in $m_{j_k}$ which are all degenerate, so they have determinant 0. Since the determinant is continuous, also the determinant of the matrix representing $\omega_\infty$ in $m_\infty$ must be 0 and thus $\omega_\infty$ is degenerate in $m_\infty$, which is against our hypotheses for $m_\infty\in Q$.
Note: $U_\delta$ is a tubular neighborhood of $Q$ of radius $\delta$, that is, we have any Riemann structure on $M$, nd we consider $N_\delta$, the subset of the normal bundle of $Q$ consisting of only vector of norm at most $\delta$, and for $\delta$ small enough the exponential map will map $N_\delta$ diffeomorphically onto an open neighborhood of $Q$, which is what we call $U_\delta$. This $U_\delta$, whenever defined, can be deformation retracted onto $Q$.
Now, apart from certain English errors (which are transcribed verbatim from those notes), and some typos (also verbatim, AFAIK), my problem in this proof is that, though $Q$ is indeed compact, $m_j$ is not a sequence in $Q$. What "one $(m_j)_{j\in\mathbb{N}}$ in $U_{1/j}$ means is that $m_j\in U_{1/j}$ for all $j$. So what has $Q$ got to do with the sequence, except for the fact that the $U_{1/j}$ are defined starting from it, and that any limit point of $(m_j)$ will have to be in $Q$ because the $U_{1/j}$ decrease towards it and so any $m_k$ is in any $U_{1/j}$ with $j\leq k$, meaning a limit point must be in all of them, hence in $Q=\bigcap_kU_{1/k}$? Or in other words:
How does compactness of $Q$ give me a converging subsequence of $(m_j)$ when $\{m_j\}_j\not\subseteq Q$?
Edit: one idea
Naturally $\{m_j\}_j\subseteq U_1$, so if one could prove that $U_1$ is (relatively) compact, then we would be done. Actually, if any of those tubular neighborhoods are relatively compact, a "tail" of the sequence (i.e. $\{m_j\}_{j\geq k$ for some $k\in\mathbb{N}$ – what's that in English? In Italian it's "coda", hence "tail") would be in a compact set and hence have a converging subsequence, and again, we'd be done. But how can I prove that they are compact? Could it be that $N_\delta$ is compact? I mean, $Q$ is, and we are limiting the norm, so perhaps $N_\delta$ is compact, and hence $U_\delta$ is compact since it's diffeomorphic (hence homeomorphic) to $N_\delta$. Is that so though?
$N_\delta$ is indeed compact.
To see this, for every $q\in Q$, consider a chart of $N$ (the whole normal bundle) centered at $(q,0)$. Let $\phi_q$ be the chart map $\phi_q:U_q\to A_q$, where $U_q\subseteq N$ is the neighborhood of $(q,0)$ where $\phi_q$ is defined, and $A_q\subseteq\mathbb{R}^k$, where $k:=\dim Q$, is the open image of $\phi_q$. For any $(q',0)\in U_q$, and any $v\in(T_{q'}Q)^\perp$, we will have $(q',v)\in U_q$ as well, because the topology in the normal bundle is a subspace topology from $TM$ and the charts there have this property. Now let's get back to $N_\delta$: we restrict $\phi_q$ to $N_\delta\cap U_q$. The image of $\phi_q$ will be of the form $A'_q\times\mathbb{R}^{d-k}$, with $d:=\dim M$, and the image of that restriction will be $A'_q\times B_{\mathbb{R}^{d-k}}(0,\delta)$. Let $\phi_q(q,0)=(a_q,0)$. Inside $A'_q\ni a_q$, we can surely find a compact set (e.g. a closed ball) containing $a_q$. So let us consider e.g. $O_q:=B_{\mathbb{R}^k}(a_q,\delta_q)\times B_{\mathbb{R}^{d-k}}(0,\delta)$, where $\delta_q$ is such that that ball is included in $A'_q$. The family $\{\phi_q^{-1}(\mathring O_q)\}$ is an open cover for the set $\{(q,0):q\in Q\}\subseteq N_\delta$, which is compact since $Q$ is, so we can find a finite subcover $\{\phi_{q_j}^{-1}(\mathring O_{q_j})\}_1^n$. That is in fact a cover of $N_\delta$ with finitely many sets, and their closures are $\phi_{q_j}^{-1}(O_{q_j})$, that is preimages via homeomorphisms of compact sets, so we have just covered $N_\delta$ with finitely many compact sets. But a finite union of compact sets is itself compact, and hence $N_\delta$ is compact.
So there you go. Problem solved.