Let $\Omega$ be a nonempty open subset of $\mathbb{R}^n$, and $\mathcal{D}(\Omega)$ the set of test functions (infinitely differentiable functions $f:\Omega \rightarrow \mathbb{C}$ with compact support contained in $\Omega$).
For every compact $K \subseteq \Omega$, let $\mathcal{D}_K$ be the locally convex topological vector space of infinitely differentiable function $f:\Omega \rightarrow \mathbb{C}$ whose support lies in $K$, with the topology $\tau_K$ induced by the system of norms ($N=0,1,2,\dots$):
\begin{equation}
\left| \left| f \right| \right|_{N} = \max \{ \left| D^{\alpha}f(x) \right| : x \in \Omega, | \alpha | =0,1,\dots, N \},
\end{equation}
where $\alpha=(\alpha_1,\dots,\alpha_n)$ is a multi-index and $|\alpha|=\alpha_1 + \dots + \alpha_n$.
The usual topology of $\mathcal{D}(\Omega)$ is defined as the strongest topology among all those topologies on $\mathcal{D}(\Omega)$ that (i) make $\mathcal{D}(\Omega)$ a locally convex topological vector space and such that (ii) the inclusion $i_K: \mathcal{D}_K \hookrightarrow \mathcal{D}(\Omega)$ is continuous for every compact $K \subseteq \Omega$. In the language of Bourbaki, $\tau$ is called the "locally convex final topology" of the family of topologies $(\tau_K)$ of the spaces $(\mathcal{D}_K)$ with respect to family of linear maps $(i_K)$.
I have two questions.
(Q1) Can we find a set $V \subseteq \mathcal{D}(\Omega)$, such that $V \cap \mathcal{D}_K \in \tau_K$ for all compact $K \subseteq \Omega$, but $V \notin \tau$?
(Q2) Can we find $V \subseteq \mathcal{D}(\Omega)$, with $0 \in V$, such that $V \cap \mathcal{D}_K \in \tau_K$ for all compact $K \subseteq \Omega$, and there is no $W \subseteq V$, with $0 \in W \in \tau$?
Clearly a positive answer to (Q2) implies that also (Q1) has a positive answer.
Note that (Q1) is equivalent to ask whether $\tau$ coincides or not with the final topology $\tau'$ on $\mathcal{D}(\Omega)$ with respect to the family of inclusions $i_K: \mathcal{D}_K \hookrightarrow \mathcal{D}(\Omega)$, where $K$ is any compact subset of $\Omega$. So we have $\tau \subseteq \tau'$ and (Q1) can maybe be given a positive indirect answer, by proving that $\tau$ and $\tau'$ do not share the same properties.
To give a positive answer to (Q2) seems to be more difficult.
Best Answer
Finally, I found the answers to my two questions, and they are both positive as I conjectured.
Take a sequence of compact sets $(K_m)_{m=0}^{\infty}$ in $\Omega$, each one with nonempty interior, such that:
(i) $K_m$ is contained in the interior of $K_{m+1}$ for each $m=0,1,\dots$;
(ii) $\cup_{m=0}^{\infty} K_m = \Omega$.
Let $(x_m)_{m=0}^{\infty}$ be a sequence in $\Omega$ such that $x_m$ lies in the interior of $K_m$ and $x_m \notin K_{m-1}$ (with $K_{-1}=\emptyset$). Define the set
\begin{equation} V = \{ f \in \mathcal{D}(\Omega) : \left| f(x_{|\alpha|}) D^{\alpha} f(x_0) \right| < 1, | \alpha |=0,1,2, \dots \}. \end{equation}
Let $K \subseteq \Omega$ be a compact set. Since only finitely many of the $x_m$'s belong to $K$, it is immediate to see that $V \cap \mathcal{D}_K \in \tau_K$. Assume that $V$ contains some $\tau$-open set containing 0. Then, since $\mathcal{D}(\Omega)$ with the topology $\tau$ is a locally convex topological vector space, there would exist a convex balanced set $W \subseteq V$ such that $W \in \tau$. So $W \cap \mathcal{D}_K \in \tau_K$ for each compact $K \subseteq \Omega$.
Then for each m, there exists a positive integer $N(m)$ and $\epsilon(m) > 0$ such that the set
\begin{equation} U_m = \left \{ f \in \mathcal{D}_{K_m}: \left| \left|f \right| \right|_{N(m)} < \epsilon(m) \right \} \end{equation}
is contained in $W \cap \mathcal{D}_{K_m}$. Let $m=N(0)+1$. Then the interior of $K_m$ contains the point $x_{N(0)+1}$, so that there exists $g \in U_m$ such that $|g(x_{N(0)+1})| > 0$. Now note that for any $M > 0$, we can find $f \in U_0$ such that $|D^{\alpha}f(x_0)| > M$ for some multi-index $\alpha$ such that $|\alpha| = N(0)+1$. This in turn implies that for any $c \in (0,1)$, we can find $f \in U_0$ such that $cf+(1-c)g$ does not belong to $V$. So $W$ is not convex, against the hypothesis. This shows that (Q2), and so (Q1), has a positive answer.
NOTE (1). Actually, this example also shows that $\mathcal{D}(\Omega)$ with the topology $\tau'$ is not even a topological vector space. Indeed, if it were, then we should be able to find $S \in \tau'$ such that $S + S \subseteq V$. Again, we could find then for each $m$ a positive integer $P(m)$ and $\delta(m) > 0$ such that the set \begin{equation} T_m = \left \{ f \in \mathcal{D}_{K_m}: \left| \left|f \right| \right|_{P(m)} < \delta(m) \right \} \end{equation} is contained in $S \cap \mathcal{D}_{K_m}$. Choose $m=P(0)+1$, so that the interior of $K_m$ contains $x_{P(0)+1}$. Then there exists $g \in T_m$ such that $|g(x_{P(0)+1})| > 0$. As before, note for any $M > 0$, we can find $f \in T_0$ such that $|D^{\alpha}f(x_0)| > M$ for some multi-index $\alpha$ such that $|\alpha| = P(0)+1$. This in turn implies that there exists $f \in T_0$ such that $f+g \notin V$. QED
NOTE (2). We can prove in the same way as before that for every $f \in V$, there is no $U \in \tau$ such that $f \in U$ and $U \subseteq V$. Assume there exists. Then, being $\mathcal{D}(\Omega)$ with the topology $\tau$ a locally convex space, we can find a convex balanced set $W \in \tau$ such that $f + W \subseteq U$. Again, we could find then for each $m$ a positive integer $N(m)$ and $\epsilon(m) > 0$ such that the set \begin{equation} U_m = \left \{ f \in \mathcal{D}_{K_m}: \left| \left|f \right| \right|_{N(m)} < \epsilon(m) \right \} \end{equation} is contained in $W \cap \mathcal{D}_{K_m}$. Choose then $m=N(0)+1$, so that the interior of $K_m$ contains the point $x_{N(0)+1}$. Then there exists $g \in U_m$ such that $|g(x_{N(0)+1})| > 0$ and $|g(x_{N(0)+1})| < | \varphi(x_{N(0)+1})|$ if $| \varphi(x_{N(0)+1})| > 0$. Now note that for any $M > 0$, we can find $f \in U_0$ such that $|D^{\alpha}f(x_0)| > M$ for some multi-index $\alpha$ such that $|\alpha| = N(0)+1$. This in turn implies that for any $c \in (0,1)$, we can find $f \in U_0$ such that $\varphi + cf+(1-c)g$ does not belong to $V$, which gives a contradiction, since $cf+(1-c)g \in W$.