It is quite well-known that locally convex inductive limits need not be Hausdorff.
Edit. As spotted by Pietro Majer, the arguments below do not work because the functions $g_{z_k}$ need not belong to a fixed $H(U_n)$ if $z\in\bigcap_{n\in\mathbb N} \overline U_n$.
There are exmples of non-Hausdorff locally convex inductive limits of nuclear Fréchet spaces in Klaus Floret's article Some aspects of the theory of locally convex inductive limits, Functional analysis: surveys and recent results, II (Proc. Second Conf. Functional Anal., Univ. Paderborn, Paderborn, 1979), pp. 205–237, North-Holland Math. Stud., 38, North-Holland, Amsterdam-New York, 1980.
Here is my favorite example. It is the (countable and injective) limit of nuclear Frechet spaces:
Let $U_n$ be a decreasing sequence of non-empty, bounded, connected, open sets in $\mathbb C$ with empty intersection and set $X_n=H(U_n)$, the space of holomorphic functions on $U_n$ endowed with the topology of uniform convergence on compact subsets ($X_n$ is thus a nuclear Frechet space). Since the restriction $X_{n}\to X_{n+1}$ $f\mapsto f|_{U_{n+1}}$ is injective we may consider $X_n$ as a subspace of $X_{n+1}$ with continuous inclusion. Then $X_\infty=$ind $X_n$ has the trivial topology.
By the Hahn-Banach theorem we have to show that every continuous linear functional $\varphi$ on $X_\infty$ is zero. To see this we define, for each $z\in\mathbb C$, the function $g_z(\omega)= 1/({z-\omega})$ which belongs to $X_n=H(U_n)$ if $z\notin U_n$. Then $h(z)=\varphi(g_z)$ defines a function on $\mathbb C$ which is holomorphic (since $(g_{z_k} - g_z)/(z_k-z)$ converges in $X_n$ if $z_k\to z$ in $U_n^c$) with $h(z_k)\to 0$ for $|z_k|\to\infty$ (since $g_{z_k}\to 0$ even in $X_1$). By Liouville's theorem, $h$ vanishes identically. Moreover, by Runge's theorem, $\{g_z: z\notin U_n\}$ has dense linear span in $X_n$ which implies that $\varphi$ vanishes on $X_n$.