Short answer: Yes. But we have to make to precise what we mean by "$X$ contains a copy of $T$". See the theorem below.
I am not sure about the relationship to Eric Wofsey's elegant solution, perhaps my answer is just a supplement.
Let $L = \{0\} \times [-1,1]$ and $S = \{(t,\sin(1/t)) \in \mathbb R^2 \mid t \in (0,1]\}$.
The closed topologist's sine curve is the space
$$T = L \cup S$$
with the subpace topology inherited from $\mathbb R^2$. It is a compactum (= compact metric space). Let us define for $r \in (0,1]$
$$S_r = T \cap \left((0,r) \times \mathbb R \right).$$
This in an open subset of $T$ homeomorphic to the open interval $(0,r)$. In fact, $S_r$ is the graph $G(\tau)$ of the function $\tau : (0,r) \to \mathbb R, \tau(t) = \sin(1/t)$.
A topologist's sine pair is a pair $(X,h)$ consisting of a topological space $X$ and a closed embedding $h : T \to X$ such that $h(S_1)$ is open in $X$. The latter condition is essential, otherwise we could take for example $X = \mathbb R^2$ and $h : T \hookrightarrow X$ which is completely uninteresting. Examples are $(T, id_T)$ and $(W,T \hookrightarrow W)$, where $W$ is the Warsaw circle $W = T \cup A$ with an arc $A \subset \mathbb R^2$ intersecting $T$ in the points $(0,-1)$ and $(1,\sin(1))$. Other examples are the covering spaces of $T$ formed by gluing $n$ copies of $T$ to a "circular" shape (for $n = 1$ we obtain $W$) and by gluing infinitely many copies of $T$ to a "line" shape.
Theorem. Let $(X,h)$ be a topologist's sine pair, $Y$ be a locally connected sequentially compact space and $f : Y \to X$ be continuous. Then there exists $r \in (0,1]$ such that $f(Y) \cap h(S_r) = \emptyset$.
Before we prove it, let us observe that it implies that $T$ is not locally connected. Just consider $id : T \to T$. The same holds for $W$.
Proof of Theorem. Assume that $f(Y) \cap h(S_r) \ne \emptyset$ for all $r \in (0,1]$. Then there exists a sequence $(y_n)$ in $Y$ such that $f(y_n) \in h(S_{1/n})$. Write $f(y_n) = h(t_n, \sin(1/t_n))$ with $t_n > 0$. Clearly $t_n \to 0$.
$(y_n)$ has a convergent subsequence; we may therefore assume w.l.o.g. that $(y_n)$ itself converges to some $y^* \in Y$. Hence $f(y_n) \to x^* = f(y^*)$. We conclude that also $x_n = h(t_n, \sin(1/t_n)) \to x^*$. Since $h(T)$ is closed in $X$, we see that $x^* \in h(T)$. Thus $(t_n,\sin(1/t_n)) \to h^{-1}(x^*) \in T$. This shows that $\xi^* = h^{-1}(x^*) \in L$, i.e. $x^* \in h(L)$. Write $\xi^* = (0,u^*)$.
The open sets $U_- = T \cap ([0,2/\pi) \times (-1,1])$ and $U_+ = T \cap ([0,2/\pi) \times [-1,1))$ cover $L$, thus one of them contains $\xi^*$. W.l.o.g. assume it is $U_+$. It is obtained from $T$ be removing the points $\xi_n = (2/(2n - 1)\pi, 1)$, $n \in \mathbb N$ and the graph $G(\tau \mid_{[2/\pi,1]})$. It is therefore the disjoint union of $U^* = U_+ \cap L$ and infinitely many open $U_n \subset S_1$ (which are the graphs $G(\tau \mid_{(2/(2n+1)\pi, 2/(2n-1)\pi)})$). The sets $U^*$ and $U_n$ are connected, but this is irrelevant here. Note that the $U_n$ are also closed in $U$. Since $h(U_+)$ is open in $h(T)$, we find an open $U' \subset X$ such that $U' \cap h(T) = h(U_+)$. Since $h(T)$ is closed in $X$, we see that $h(U_+)$ is closed in $U'$. Hence all $h(U_n)$ are closed in $U'$. But $h(S_1)$ is open in $X$, thus the $h(U_n)$ are open subsets of $X$ which are contained in $U'$, thus they are open in $U'$. Hence the $h(U_n)$ are clopen in $U'$.
Let $V$ be a connected open neigborhood of $y^*$ such that $f(V) \subset U$. We have $y_n \in V$ for $n \ge n_0$. Since $f(V)$ is connected, we conclude that if $f(V)$ would intersect one of the clopen sets $h(U_n)$, then $f(V) \subset h(U_n)$. But $x^* = f(y^*) \in f(V)$, thus $f(V) \subset h(U_n)$ is impossible. Hence $f(V) \cap h(U_n) = \emptyset$ for all $n$. On the other hand $x^* \in h(U_+) \subset h(T)$, thus $f(y_n) \to x^*$ implies that $f(y_n) \in h(U_+)$ for $n \ge n_1 \ge n_0$. In other words, $f(V)$ intersects $h(\bigcup_n U_n) = \bigcup_n h(U_n)$ because by construction $f(y_n) \notin h(U^*)$. This contradiction proves the theorem.
