In Conway's Functional Analysis textbook, exercise-IV.2.8 asks to prove the open mapping theorem for Frechet spaces. In Conway's book, a Frechet space is not required to be locally convex.
Let $X,Y$ be Frechet spaces, and $T: X\to Y$ a surjective continuous linear map. Then $T$ is an open map.
I can mimic the proof for Banach spaces but get stuck after the step where the image of a particular open ball contains an open ball in $Y$.
Attempted Proof. Since, in metric spaces, the open balls are basic open sets, it suffices to show, for every $r>0$ and $x\in X$, $T(B_X(x, r))$ is open in $Y$. By the homeomorphism of translation, it then suffices to show $T(B_X(0, r))\ni 0_Y$ is open in $Y$. Since $T$ is surjective, we have $$Y=T(X)=T(\bigcup_{n\in\mathbb{N}} B_X(0, nr))=\bigcup_{n\in\mathbb{N}}T(B_X(0, nr)).$$ Since Frechet spaces are complete metric spaces, by Baire Category Theorem, there exists $n_0\in\mathbb{N}$, such that $\text{Int}\overline{T(B_X(0, n_0r))}\neq \emptyset,$ and so there exist $y_0\in Y$ and $r_0>0$ such that $$B_Y(y_0, r_0)\subset \overline{T(B_X(0, n_0r))}\Rightarrow y_0, y_0+v \in \overline{T(B_X(0, n_0r))},$$ where $v\in B_Y(0, r_0)$. Hence, for every such $v$, we have $$v\in \overline{T(B_X(0, 2n_0r))}\Rightarrow B_Y(0, r_0)\subset \overline{T(B_X(0, 2n_0r))}.$$
Next, we must show there are no holes in $T(B_X(0, 2n_0r))\cap B_Y(0, r_0)$. Same as with Banach spaces cases, we can do so by finding a series $\sum_n x_n$, whose image under $T$ converges to arbitrary point $y \in B_Y(0, r_0)$.
Note, for every $m\in\mathbb{N}$, there exists $r_{0,m}>0$ such that $B_Y(0, r_{0,m})\subset \overline{T(B_X(0, \frac{2n_0r}{2^m}))}.$ This is true because we can repeat the Baire Category argument as above for every radius $\frac{2n_0r}{2^m}$. Define $\tilde{r_{0,m}}:=\min\{r_{0,m}, \frac{r_0}{2^m}\}$. Then, we have $$B_Y(0, \tilde{r_{0,m}})\subset \overline{T(B_X(0, \frac{2n_0r}{2^m}))}.$$ Therefore, if we start from an arbitary point $y_1\in B_Y(0, r_0)$, there exist $x_1\in B_X(0, 2n_0r)$, and by induction, we can have two sequences $(x_m)$ and $(y_m)$ such that $$x_m\in B_X(0, \frac{2n_0r}{2^m}); \quad y_{m+1}=y_m-Tx_m \in B_Y(0, \tilde{r_{0,m}})\subset B_Y(0, \frac{r_{0}}{2^m}).$$ By completeness of $X$, the series $\sum_m x_m$ then converges absolutely to some $\xi\in B_X(0, 4n_0r)$, and so $$y_{m+1}=y_m -Tx_m =y_1 -T(\sum_{i=1}^mx_i) \to 0\Rightarrow y_1=T\xi\Rightarrow B_Y(0, r_0)\subset T(B_X(0, 4n_0r)).$$
However, I don't know how to use what I've proved so far to finally show it also works for every open ball, since in Frechet spaces, the metric is not homogeneous. So we can't simply multiply an arbitary scalar to make a fixed open ball cover any open ball with a different radius.
Update: After consulting Rudin's textbook, it seems Baire category theorem should be applied at the following decomposition: $$Y=T(X)=T(\bigcup_{n\in\mathbb{N}} n\cdot B_X(0, r))=\bigcup_{n\in\mathbb{N}}n\cdot T(B_X(0, r)).$$ Then, there exists $n_0\in\mathbb{N}$, $y_0\in Y$ and $r_0>0$ such that $$B_Y(y_0, r_0)\subset \overline{n_0\cdot T(B_X(0, r))}=n_0\cdot\overline{ T(B_X(0, r))}.$$ Then, there exists a $r_0'\in (0, \frac{r_0}{n_0})$ such that $$B_Y(y_0, r_0')\subset \overline{ T(B_X(0, r))}.$$ Everything can then proceed from there without obstacles.