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The answer is no. Take the "classical" example of the line with two origins. This space is non-Hausdorff, paracompact and doesn't admit partitions of unity.

EDTI: I think the question is a kind of "duplicate" . Ok, but if you have an example for a non-Hausdorff manifold, which doesn't admit partitions of unity, you have an example for a non-Hausdorff paracompact space with the same property.

First the definition:
The line with two origins is the quotient space of two copies of the real line $\mathbb{R} \times {a}$ and $\mathbb{R} \times {b}$.
with equivalence relation given by $(x,a) \sim (x,b)\text{ if }x \neq 0$.
Since all neighbourhoods of $0_a$ intersect all neighbourhoods of $0_b$, it is non-Hausdorff.
However, this space is paracompact, since $\mathbb{R}$ is paracompact.

For the non-existence of a partition of unitiy: take the open covering $ U = (-\infty,0) \cup \{ 0_a \} \cup (0,\infty)$ and $\tilde{U} = (-\infty,0) \cup \{ 0_b \} \cup (0,\infty)$. Assume, there is a partition of unity subordinate to this cover. Then the value of each origin would have to be $1$ which cannot be true. (Edit: villemoes was a little faster :-) )

The answer is no. Take the "classical" example of the line with two origins. This space is non-Hausdorff, paracompact and doesn't admit partitions of unity.

EDIT: I think the question is a kind of "duplicate" . Ok, but if you have an example for a non-Hausdorff manifold, which doesn't admit partitions of unity, you have an example for a non-Hausdorff paracompact space with the same property.

First the definition:
The line with two origins is the quotient space of two copies of the real line $\mathbb{R} \times {a}$ and $\mathbb{R} \times {b}$.
with equivalence relation given by $(x,a) \sim (x,b)\text{ if }x \neq 0$.
Since all neighbourhoods of $0_a$ intersect all neighbourhoods of $0_b$, it is non-Hausdorff.
However, this space is paracompact, since $\mathbb{R}$ is paracompact.

For the non-existence of a partition of unity: take the open covering $ U = (-\infty,0) \cup \{ 0_a \} \cup (0,\infty)$ and $\tilde{U} = (-\infty,0) \cup \{ 0_b \} \cup (0,\infty)$. Assume, there is a partition of unity subordinate to this cover. Then the value of each origin would have to be $1$ which cannot be true. (Edit: villemoes was a little faster :-) )

The answer is no. Take the "classical" example of the line with two origins. This space is non-Hausdorff, paracompact and doesn't admit partitions of unity.

EDTI: I think the question is a kind of "duplicate" . Ok, but if you have an example for a non-Hausdorff manifold, which doesn't admit partitions of unity, you have an example for a non-Hausdorff paracompact space with the same property.

First the definition:
The line with two origins is the quotient space of two copies of the real line $\mathbb{R} \times {a}$ and $\mathbb{R} \times {b}$.
with equivalence relation given by $(x,a) \sim (x,b)\text{ if }x \neq 0$.
Since all neighbourhoods of $0_a$ intersect all neighbourhoods of $0_b$, it is non-Hausdorff.
However, this space is paracompact, since $\mathbb{R}$ is paracompact.

For the non-existence of a partition of unitiy: take the open covering $ U = (-\infty,0) \cup \{ 0_a \} \cup (0,\infty)$ and $\tilde{U} = (-\infty,0) \cup \{ 0_b \} \cup (0,\infty)$. Assume, there is a partition of unity subordinate to this cover. Then the value of each origin would have to be $1$ which cannot be true. (Edit: villemoes was a little faster :-) )

The answer is no. Take the "classical" example of the line with two origins. This space is non-Hausdorff, paracompact and doesn't admit partitions of unity.

EDTI: I think the question is a kind of "duplicate" . Ok, but if you have an example for a non-Hausdorff manifold, which doesn't admit partitions of unity, you have an example for a non-Hausdorff paracompact space with the same property.

First the definition:
The line with two origins is the quotient space of two copies of the real line $\mathbb{R} \times {a}$ and $\mathbb{R} \times {b}$.
with equivalence relation given by $(x,a) \sim (x,b)\text{ if }x \neq 0$.
Since all neighbourhoods of $0_a$ intersect all neighbourhoods of $0_b$, it is non-Hausdorff.
However, this space is paracompact, since $\mathbb{R}$ is paracompact.

For the non-existence of a partition of unitiy: take the open covering $ U = (-\infty,0) \cup \{ 0_a \} \cup (0,\infty)$ and $\tilde{U} = (-\infty,0) \cup \{ 0_b \} \cup (0,\infty)$. Assume, there is a partition of unity subordinate to this cover. Then the value of each origin would have to be $1$ which cannot be true. (Edit: villemoes was a little faster :-) )

The answer is no. Take the "classical" example of the line with two origins. This space is non-Hausdorff, paracompact and doesn't admit partitions of unity.

EDTI: I think the question is a kind of "duplicate":

Smooth manifolds that don't admit a partition of unity

The answer is no. Take the "classical" example of the line with two origins. This space is non-Hausdorff, paracompact and doesn't admit partitions of unity.

EDTI: I think the question is a kind of "duplicate" . Ok, but if you have an example for a non-Hausdorff manifold, which doesn't admit partitions of unity, you have an example for a non-Hausdorff paracompact space with the same property.

First the definition:
The line with two origins is the quotient space of two copies of the real line $\mathbb{R} \times {a}$ and $\mathbb{R} \times {b}$.
with equivalence relation given by $(x,a) \sim (x,b)\text{ if }x \neq 0$.
Since all neighbourhoods of $0_a$ intersect all neighbourhoods of $0_b$, it is non-Hausdorff.
However, this space is paracompact, since $\mathbb{R}$ is paracompact.

For the non-existence of a partition of unitiy: take the open covering $ U = (-\infty,0) \cup \{ 0_a \} \cup (0,\infty)$ and $\tilde{U} = (-\infty,0) \cup \{ 0_b \} \cup (0,\infty)$. Assume, there is a partition of unity subordinate to this cover. Then the value of each origin would have to be $1$ which cannot be true. (Edit: villemoes was a little faster :-) )